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\title[MATH 234 Spring 14]{Vanderbilt University, MATH 234 Spring 14: Some preparatory material}
%\date{\today}

%\author[Disconzi]{Marcelo M. Disconzi}
%\address{Department of Mathematics\\
%Vanderbilt University, Nashville, TN 37240, USA}
%\email{marcelo.disconzi@vanderbilt.edu}
\urladdr{http://www.disconzi.net/Teaching/MAT234-Spring-14/MAT234-Spring-14.html}

%\date{\today}
%\pagestyle{empty}

\begin{document}


%\begin{abstract}
%Abstract. 
%\end{abstract}


\maketitle

\tableofcontents

\section{Introduction.}

The purpose of these notes is to review some basic notions, set up the notation,
and give students an idea of some of the more basic material that will be required
for the course. While not all of this material is necessarily a pre-requisite,
it is something  that students are expected to quickly acquaint themselves with.
Some students may find most of what follows very basic. If you feel that way, 
you are still urged to go over these notes carefully and make sure you understand them, 
as some of its aspects contain subtleties that are many times overlooked when one
first learns the material.

While we tried to give a sufficiently precise treatment of the concepts involved,
our approach is primarily \emph{pragmatic}, invoking certain mathematical ideas only
as long as they are necessary to solving partial differential equations in $\RR^n$. As a result,
 we have avoided
the level of rigor usually employed when many of these ideas are first introduced
to a mathematical audience.

The conventions and notation described below will be adopted throughout the course, unless stated otherwise. While for the most part, they  are compatible with those of the course 
textbook \cite{PR},
there are some differences, so please be alert.

\section{Vectors and coordinates.\label{vect_coord}}

Recall that $\RR^n$ consists of the set of $n$-tuples $(x_1,x_2, \dots, x_n)$, where
each $x_i$, $i=1,2,\dots,n$, is a real number. This last statement is written
briefly as $x_i \in \RR$, where $\RR$ denotes the set of real numbers and $\in$ means
``belongs to" (the symbol $\notin$ will be used later on, and it means ``does not belong to").
 We write $x \in \RR^n$ to indicate that $x$ is an element of $\RR^n$, i.e.,
$x = (x_1,x_2, \dots, x_n)$. $n$ is called the \textbf{dimension} of the space
$\RR^n$. Each $x_i$ is called a \textbf{component}
or \textbf{coordinate} of $x$ ($x_1$ is the first component or first coordinate, 
$x_2$ the second component or coordinate, and so on). We shall use the terminology 
\textbf{point} and \textbf{vector} interchangeably for elements of $\RR^n$. 
``Component" and ``coordinate" will also be used interchangeably.
Notice that we \emph{do not} use an ``arrow," i.e., 
 the notation $\vec{x}$ (that you might have seen 
in Physics, Linear Algebra, or Multi-variable Calculus) for elements of $\RR^n$.
In particular, the \textbf{zero element} of $\RR^n$, also called
the \textbf{origin},  will be denoted simply by $0$, i.e.
$0 \in \RR^n$ corresponds to the vector whose components are all zero:
$0 = (0,0,\dots, 0)$.

While we typically use letters such as $x, y$, and $z$ to denote elements 
of $\RR^n$, sometimes, when dealing with $\RR^2$ and $\RR^3$, we reserve them
for the components of a vector. For example, $v = (x,y,z) \in \RR^3$.
On the other hand, in many situations we have to label a \emph{set of points} in 
$\RR^n$, in which case we use a subscript. For example, suppose we are given 
$N$ points\footnote{Note that $N$ and $n$ have nothing to do with each other. $N$
denotes the number of points  we are given or have chosen, so if we pick, say, ten points,
we have $N=10$. $n$ denotes the dimension of the space. In a given problem, $n$
is usually fixed (say, we are working on the three-dimensional space $\RR^3$), 
while $N$ can vary (first we are given ten points, $N=10$; later on we pick six points, $N=6$, 
etc.).\label{footnote_N_n}} in $\RR^n$. 
We can denote them by $x_1, x_2, \dots, x_N$. Here,
we have $x_i \in \RR^n$ for each $i=1, 2, \dots, N$. In this case $x_i$
\emph{should not be confused} with the $i^\text{th}$ component of a vector, being
rather the $i^\text{th}$ vector in the collection of $N$ vectors
$x_1, x_2, \dots, x_n$. Naturally, in such a situation a different notation
is needed for the components of the vectors $x_1, x_2, \dots, x_N$. For example, 
if we want to write the components of, say, the vector $x_3$, we can use two indices, as
\begin{gather}
x_3 = (x_{31}, x_{32}, \dots, x_{3n}).
\nonumber
\end{gather}
Above, the first index labels which vector we are talking about, i.e., the third vector 
in this case, whereas the second index denotes the corresponding component of the third vector.
More generally, we write,
\begin{gather}
x_i = (x_{i1}, x_{i2}, \dots, x_{in}),
\nonumber
\end{gather}
where $i$ indicates the $i^\text{th}$ vector in the collection 
$x_1, x_2, \dots, x_N$, and the second index indicates the corresponding
component of the $i^\text{th}$ vector. Therefore, we can denote all components
of all the vectors in our collection by $x_{ij}$, where $i$ varies from $1$ to $N$ 
and labels which vector we are talking about, and $j$ varies from $1$ to $n$ and denotes
the $j^\text{th}$ component of the $i^\text{th}$ vector.

Yet sometimes, it will be more convenient to denote the coordinates of a point
with \emph{upper indices} or \emph{super-scripts}, i.e.,
\begin{gather}
x  = (x^1, x^2, \dots, x^n),
\nonumber
\end{gather}
in which case the components of a collection of $N$ vectors can be written 
as  $x^j_i$, with $i$ and $j$ holding the same interpretation as in the last paragraph, i.e., 
$i$ varies from $1$ to $N$ 
and labels which vector we are talking about, and $j$ varies from $1$ to $n$ and denotes
the $j^\text{th}$ component of the $i^\text{th}$ 
vector\footnote{Although we shall not explore it in this course, there is in fact
a deeper meaning in the distinction between denoting coordinates as $x^j$ and $x_j$.
The mathematically inclined reader can search for the concept of the \emph{dual of a vector space},
whereas Physics students may want to look at the distinction between \emph{vectors} and 
\emph{co-vectors},
or  between \emph{covariant} and \emph{contravariant} coordinates.}.

All of the above may look very confusing at first sight. Sometimes $x_i$ is 
a component, sometimes it denotes a point (=vector!) in $\RR^n$; but sometimes
components are denoted by $x$ and $y$. The point, of course, is that it may be
more convenient to use one notation over the other. It all depends
on the particular problem we are addressing, and we need to have enough flexibility 
to use the most convenient notation in each case.

The important thing to have in mind is that what determines which type of
object one has is not its notation, but rather how it is stated
in the relevant context. For instance, suppose you are given a homework
problem that begins with ``Let $x \in \RR^n$...". In this case, you are told
that $x$ is a vector with $n$ components. Similarly, you could find a statement
in the textbook that reads ``Consider three points, $x_1$, $x_2$, $x_3$, in $\RR^3$...".
Here, you are told that each $x_i$, $i=1,2,3$, is a three-component vector\footnote{Notice that 
it is just a coincidence that the number $3$ appears twice here: we could have more, or less,
than three points in the three-dimensional space $\RR^3$; see footnote \ref{footnote_N_n}.}.
Similarly, whenever you write your solutions, you should  make clear what  you mean
by each object you introduce. Except in very special cases, you 
should not, and cannot, assume that it is obvious what is meant by the notation you are using.
For example, if you are given a problem that says ``Let $x \in \RR^n \dots$", then it will
be clear that by $x_i$ you mean the components of $x$. On the other hand, if the problem
makes no reference to $x$, then you cannot write things like $x_i$ assuming that
the reader will know what you mean, as it could be interpreted in several different ways.
In this case, you have to write in your solutions something like ``Let $x \in \RR^n$...",
or whatever else is meant in the context at hand.

While notation can, in fact, be a source of a great deal of confusion, it is important
to know how make the most of the convenience that 
different notations offer. Naturally, the importance resides in the concepts themselves
rather than in how one expresses them, although having a clear way of expressing 
mathematical ideas is definitely preferable. This is not, of course, much 
different than when you first learned about variables and functions, probably 
sticking to writing $x$ for the variable and $y$ for the function, learning, later on,
that you could use different letters without changing the mathematical content
of the problems.

The \textbf{canonical} or \textbf{standard} vectors in $\RR^n$, denoted
by $e_i$, $i=1, \dots, n$, are the vectors with $1$ in the $i^\text{ih}$ 
component and zero in the remaining ones, i.e.,
\begin{align}
\begin{split}
e_1 & = (1, 0, 0, \dots, 0 ) \\
e_2 & = (0, 1, 0, \dots, 0) \\
e_3 & = (0, 0, 1, \dots, 0) \\
\vdots & \hspace{1.5cm} \vdots \\
e_n & = (0, 0, 0, \dots, 1 ).
\end{split}
\nonumber
\end{align}
We can write the above also as
\begin{align}
\begin{split}
e_i & = (0,0, \dots, 1, 0, 0, \dots, 0) \\
& \hspace{2cm} \uparrow \\
& \hspace{1cm}i^\text{th} \text{ component.}
\end{split}
\nonumber
\end{align}
According to the conventions previously discussed, the components of $e_i$ can be described as
\begin{gather}
e_{ij} = 
\begin{cases}
1, & \text{ if } i = j,\\
0, & \text{ if } i \neq j.
\end{cases}
\nonumber
\end{gather}
The \textbf{norm} of a vector $x \in \RR^n$, denoted by $|x|$, is defined by
\begin{gather}
|x| = \sqrt{ x_1^2 + x_2^2 + \cdots + x_n^2 }.
\nonumber
\end{gather}
Notice that $|x|$ measures the distance of $x$ to the origin, or, equivalently,
the ``length" of the vector $x$. If $x,y \in \RR^n$, then $|x-y|$ is simply the distance
between $x$ and $y$.

The \textbf{inner product} or \textbf{dot product} between two vectors $x$ and $y$ of
$\RR^n$ is denoted by both $\langle x, y \rangle$ and $x \cdot y$, and it is defined as
\begin{gather}
\langle x, y \rangle = x_1 y_1 + x_2 y_2 + \cdots x_n y_n,
\nonumber
\end{gather}
or, more concisely,
\begin{gather}
\langle x, y \rangle = \sum_{i=1}^n x_i y_i.
\nonumber
\end{gather}
Notice that $|x| = \sqrt{ \langle x, x \rangle }$. Two vectors are said to be
\textbf{orthogonal} if their inner product is zero.

\section{Functions.\label{functions}}
We recall that a \textbf{function} is a rule that assigns for each element in a set
$A$, one, and only one, element in a set $B$. The set $A$ is called the 
\textbf{domain} of the function and $B$ its \textbf{co-domain}. The notation
\begin{gather}
f: A \rar B
\nonumber
\end{gather}
is used to indicate that $f$ is a function with domain $A$ and co-domain $B$.
We sometimes say that $f$ \textbf{takes values in} $B$ to refer to the co-domain of a function.

For the most part, we shall be dealing with functions defined in $\RR^n$ (or in a subset
of $\RR^n$, see section \ref{sets}) 
and taking values in $\RR$, i.e., $f: \RR^n \rar \RR.$ In this case,
if we write $f(x)$, it is to be understood that $x \in \RR^n$, i.e., $x = (x_1,x_2,
\dots, x_n)$ and $f(x) = f(x_1,x_2,\dots, x_n)$. A function that takes values in $\RR$
is called a \textbf{real valued function}. The following are examples of real valued
functions:
\begin{align}
\begin{split}
(i) 
&
\begin{array}{ll}
 & f: \RR \rar \RR, \\
  & f(x) = 2x + 1. 
\end{array}
\\
(ii)
&
\begin{array}{ll}
 & g: \RR^3 \rar \RR, \\
& g(x,y,z) = xy + xz + yz.
\end{array}
\\
(iii)
&
\begin{array}{ll}
 & h: \RR^n \rar \RR, \\
& h(x) = \langle x, x \rangle. 
\end{array}
\end{split}
\nonumber
\end{align}
The above are typical examples of how a function is usually presented. Consider example (ii).
First, we are
given something like $g: A \rar B$, which indicates how we are naming the function ($g$ in this case),
and the domain and co-domain of the function ($\RR^3$ and $\RR$, respectively); after all, 
if a function is a rule between two sets, upon defining a function we better say  what these
two sets are. Next, we define the rule that associates to each element in $A$ one, and only one, element 
in $B$. In this example, the rule is the following: given an element $(x,y,z) \in \RR^3$,
the corresponding element in $\RR$ is obtained by computing $xy + xz + yz$. For example,
$g(1,-2,3) = 1\times (-2) + 1 \times 3 + (-2)\times 3 = -5$. Although this all sounds very trivial,
you should make sure that you understand the notation involved.

Less often, we shall need functions with domain $\RR^n$ (or a subset of $\RR^n$, see 
section \ref{sets}) 
and $\RR^m$, where $n$ and $m$ may or may not be equal, depending on the particular problem.
A function $f: \RR^n \rar \RR^m$ is called a \textbf{vector valued function}.
The particular case when $n=m$ is called a \textbf{vector field}.
The following are examples:
\begin{align}
\begin{split}
(iv) 
&
\begin{array}{ll}
 & u: \RR^2 \rar \RR^3, \\
  & u(x) = (x_1 x_2, x_1 - x_2, x_1 + x_2). 
\end{array}
\\
(v)
&
\begin{array}{ll}
 & v: \RR^2 \rar \RR^2, \\
& v(x,y) = (\frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{2} y, 
\frac{\sqrt{2}}{2} x + \frac{\sqrt{2}}{2} y )
\end{array}
\\
(vi)
&
\begin{array}{ll}
 & w: \RR^n \rar \RR^n, \\
& w(x) = -x. 
\end{array}
\end{split}
\nonumber
\end{align}
Notice the difference between examples $(i)-(iii)$ and $(iv)-(vi)$. In 
$(i)-(iii)$ the answer is a \emph{real number}, while in $(iv)-(vi)$
the answer is always a \emph{vector} with more than one component. For instance,
given $(1,1) \in \RR^2$, $u(1,1)$ is the vector $(1,0,2) \in \RR^3$. 

We remark that sometimes functions have a nice geometric interpretation. 
You are invited to explore the geometric meaning of the function in example 
$(v)$. By assigning 
values to $x$ and $y$ and drawing both $(x,y)$ and $v(x,y)$, you should be able
to verify that $v$ rotates vectors in $\RR^2$ by $45^\circ$ counter-clockwise.
The function $w$ in example $(vi)$ is a ``reflection": it sends each point $x\in \RR^n$
to its antipodal point.

Looking at $(iv)-(vi)$, we see that a vector valued function can be thought of as 
a vector where \emph{each component is a real-valued function}. Thus, we can write,
\begin{align}
\begin{split}
& f: \RR^n \rar \RR^m,\\
& f = (f_1, f_2, \dots, f_m),
\end{split}
\nonumber
\end{align}
where
\begin{gather}
f_i: \RR^n \rar \RR, \, i = 1,\dots, m.
\nonumber
\end{gather}
Even more explicitly, since each $f_i$ has domain $\RR^n$, we can write,
\begin{align}
\begin{split}
& f: \RR^n \rar \RR^m,\\
& f(x) = (f_1(x), f_2(x), \dots, f_m(x)), \\
& \hspace{0.75cm} = (f_1(x_1, x_2, \dots,x_n), f_2(x_1, x_2, \dots, x_n), \dots, f_m(x_1,x_2, \dots, x_n) ).
\end{split}
\nonumber
\end{align}
Notice that it is not always true that a ``rule between sets" gives a well-defined function. 
Consider
\begin{align}
\begin{split}
& f: \RR^4 \rar \RR,\\
& f(x) = x.
\end{split}
\nonumber
\end{align}
While $f(x) = x$ seems a perfectly well-defined rule, it is inconsistent with 
$f: \RR^4 \rar \RR$, in that this last statement says that the co-domain is $\RR$, but 
$x \in \RR^4$. We can correct this by either changing the co-domain to $\RR^4$, or by
changing the form of $f$ to make it real valued, for example, $f(x) = |x|$.

Consider a function $f:A \rar B$. Notice that not all elements of $B$ have to be ``hit"
by the function. For instance, we can define $f:\RR \rar \RR$ by $f(x) = x^2$. Then
the co-domain is the set of all real numbers, but if we pick $-1 \in \RR$, there is no
$x$ in the domain such that $f(x) = -1$. Given an element $y \in B$, we say that it is the 
\textbf{image} (through $f$)
of the element $x \in A$ if $f(x) = y$. The \textbf{range} (sometimes
also called the \textbf{image set} or just image for short) of a function $f: A \rar B$
is the set of elements in $B$ that are the image through $f$ of at least one element 
from $A$. In the previous example, the range of $f(x) = x^2$ is the set of non-negative
real numbers.

\section{Sets and subsets, and more about functions.\label{sets}}
In many situations it will be necessary to work with functions
defined on subsets of $\RR^n$. We shall also need to look at subsets of points and 
functions
with certain properties, and therefore, we  introduce here some general notions of
how to carry out such definitions.

A common way to define a set of elements with a certain property is to use 
\textbf{curly brackets to mean ``set"} and  \textbf{the symbol $\Big |$ to mean
``such that"}. For example, consider the set $Q$ defined by
\begin{gather}
Q = \Big \{ x \in \RR^2 \, \Big | \, x_2 \geq 0 \Big \}.
\nonumber
\end{gather}
One reads this as ``Q is the set of all points in $\RR^2$ such that the second 
coordinate is non-negative." ``All points in $\RR^2$" is indicated by 
$x \in \RR^2$; ``such that" is indicated by $\Big |$; and ``the second coordinate is non-negative"
is indicated by $x_2 \geq 0$. In other words, $Q$ consists of the first and second quadrants
together.

We can have more than one property defining a set. For example,
\begin{gather}
I = \Big \{ x \in \RR^2 \, \Big | \, x_1 \geq 0, \, \text{ and }\, x_2 \geq 0 \Big \}
\nonumber
\end{gather}
consists of all points in $\RR^2$ such that the first coordinate is non-negative 
\emph{and} the second coordinate is non-negative as well. In other words, $I$ is the 
first quadrant on the plane. Most of the time we omit the conjunction ``and," listing
the defining properties of a set separated simply by a comma. Thus, the set $I$
above could also be defined as
\begin{gather}
I = \Big \{ x \in \RR^2 \, \Big | \, x_1 \geq 0,  x_2 \geq 0 \Big \}.
\nonumber
\end{gather}

Notice that above a certain convention for how to denote coordinates in $\RR^2$ is implicitly
understood (see the discussion on section \ref{vect_coord}). In most cases we assume
that the notation for the elements defining a set speaks for itself. For instance,
the reader should have no trouble in understanding that
\begin{gather}
S^1= \Big \{ (x,y) \in \RR^2 \, \Big | \, x^2 + y^2 = 1 \Big \}
\nonumber
\end{gather}
defines a circle of radius one centered at the origin, with $x$ and $y$ denoting
the first and second coordinates in $\RR^2$, respectively.
Another example is
\begin{gather}
B^1= \Big \{ (x,y) \in \RR^2 \, \Big | \, x^2 + y^2 \leq 1 \Big \}.
\nonumber
\end{gather}
The set $B^1$ is a ``ball" of radius one centered at the origin, i.e., not only
the circle of radius one but also its interior. Yet another simple example is
\begin{gather}
A = \Big \{ (x,y) \in \RR^2 \, \Big | \, x^2 + y^2 \leq 1, \, x^2 + y^2 \geq \frac{1}{4} \Big \}.
\nonumber
\end{gather}
How does $A$ look like? Recall that $x^2 + y^2$ is the \emph{square} of the distance from the origin
to $(x,y)$. Thus,  it reads, ``$A$ is the set of all points
in $\RR^2$ such that their distance to the origin is less than or equal to one, \emph{and}
their distance to the origin is greater than or equal to $\frac{1}{2}$." We conclude that 
$A$ is the annular region between the circle of radius one and the circle of radius $\frac{1}{2}$. 
Notice that $A$ could also have been written as
\begin{gather}
A = \Big \{ (x,y) \in \RR^2 \, \Big | \, \frac{1}{4} \leq  x^2 + y^2 \leq 1 \Big \}.
\nonumber
\end{gather}
The definition of a set can also carry a \emph{parameter}. Consider
\begin{gather}
S_r = \Big \{ (x,y) \in \RR^2 \, \Big | \, x^2 + y^2 = r^2 \Big \}.
\nonumber
\end{gather}
Clearly, $S_r$ consists of the circle of radius $r$ centered at the origin. Notice that $r$ does not 
appear together with $(x,y) \in \RR^2$, i.e., we are not saying that $S_r$ is ``the set of 
$r$'s with a certain property." Rather, \emph{given $r$}, we define $S_r$ so that
 for each $r$ we pick, there
is a different set $S_r$. In other words, $r$ is something we typically fix first, depending
on the nature of the problem we want to deal with, and then we define the set $S_r$.

Consider this other example:
\begin{gather}
S_r(x_0, y_0) = \Big \{ (x,y) \in \RR^2 \, \Big | \, (x-x_0)^2 + (y-y_0)^2 = r^2 \Big \}.
\nonumber
\end{gather}
$S_r(x_0, y_0)$ is the circle of radius $r$ centered at $(x_0,y_0)$. Again, notice that 
$(x_0,y_0)$ is something \emph{fixed} for the definition of the set, i.e., we are not considering
``all $(x_0,y_0)$ in $\RR^2$ such that...". A slightly more elaborate example is
\begin{gather}
\Pi_y = \Big \{ x \in \RR^n \, \Big | \, \langle x, y \rangle = 0 \Big \}.
\nonumber
\end{gather}
$\Pi_y$ is the ``set of all vectors in $\RR^n$ such that their inner product with 
the (fixed) element $y$ is equal to zero." In other words, $\Pi_y$ is the set of all
vectors in $\RR^n$ that are orthogonal to $y$ (the reader familiar with Linear Algebra will
recognize $\Pi_y$ as a $n-1$-dimensional place through the origin in $\RR^n$). Notice that, similarly
to $r$ above,  
 the element $y$ is fixed beforehand.
 
Now that we know how to define sets in $\RR^n$, we can look at functions whose domain is not the whole
of $\RR^n$ but a subset of it. Consider
\begin{gather}
B_1^n= \Big \{ x \in \RR^n \, \Big | \, |x| < 1 \Big \}.
\nonumber
\end{gather}
$B_1^n$ is ``the ball of radius one inside $\RR^n$," i.e., the set of all vectors in 
$\RR^n$ whose norm is less than  one. Define
\begin{align}
\begin{split}
& f: B_1^n \rar \RR,\\
& f(x) = |x|.
\end{split}
\nonumber
\end{align}
By definition, this function has domain  $B_1^n$, and thus it does not make sense
to ask what $f(x)$ is if $ x \notin B_1^n$ (remember that $\notin$ means ``does not belong to").
Some readers may find this example a little silly, as we can perfectly compute $|x|$ for any
$x \in \RR^n$, and not only for those in $B_1^n$. However, such readers should remember that
a function is a rule between two sets, $A$ and $B$, and as such it does not make sense
to ask what should be assigned to elements outside\footnote{The fact that the same formula, 
$f(x) = |x|$, holds for elements outside $B_1^n$ means that our function $f$, initially defined
on $B_1^n$, can be \emph{extended} to the whole of $\RR^n$ and, moreover, this extension
is in a sense ``the obvious" and ``best" one. We shall not get into this type of conceptual 
subtlety in our course, but the mathematically inclined students is welcome to discuss
this with me during office hours.} $A$. If this puzzles you, you can consider that this is a
``minimalist" approach: if in a particular problem or application we only care about
points in $B_1^n$, we restrict ourselves to defining functions on in $B_1^n$, not 
caring about what happens outside $B_1^n$.

While for some readers the discussion of the last paragraph may sound conceptually
abstract, there are of course obvious cases where the defining rule of a function 
only holds in a subset of $\RR^n$. E.g., let
\begin{gather}
\RR^n_0 = \Big \{ x \in \RR^n \, \Big | \, x \neq 0 \Big \},
\nonumber
\end{gather}
i.e., $\RR^n_0$ is $\RR^n$ except for the origin. Then
\begin{align}
\begin{split}
& f: \RR^n_0 \rar \RR,\\
& f(x) = \frac{1}{|x|}
\end{split}
\nonumber
\end{align}
is a well defined function, but the same expression cannot be used on the whole of $\RR^n$
as we would otherwise divide by zero. The set $\RR^n_0$ is more commonly written as
\begin{gather}
\RR^n_0 = \RR^n \backslash \{ 0 \},
\nonumber
\end{gather}
where $\backslash$ means \textbf{minus}, i.e, $\RR^n_0$ is the set of all elements in $\RR^n$
minus (i.e., except) the set consisting only of the element zero, $\{ 0 \}$.

Another example is
\begin{align}
\begin{split}
& f: \Big \{ x \in \RR^n \, \Big | \, |x| \leq 1 \Big \} \rar \RR,\\
& f(x) = \sqrt{ 1 - |x|}.
\end{split}
\nonumber
\end{align}
$f$ will not be real valued unless we restrict it to $|x| \leq 1$. 
In this last example, we took a shortcut: instead of first defining 
the set $\Big \{ x \in \RR^n \, \Big | \, |x| \leq 1 \Big \}$, and then defining $f$ as
having that set as domain,
we preferred to define $f$ and its domain simultaneously.
 This is useful to avoid ``giving names" to several different 
sets every time we define a new function.

We finish, noticing that there is a natural way to define $f(U)$, where $U$ is
some subset of the domain of $f$. If, say, $f: \RR^n \rar \RR$ and $U\subset 
\RR^n$, then
\begin{gather}
f(U) = \Big \{ f(x) \, \Big | \, x \in U \Big \}.
\nonumber
\end{gather}
In particular, if $f:A \rar B$, then $f(A)$ is simply the range of $f$.

\section{Some topological notions\footnote{In this section we try to introduce some ideas from Topology 
without assuming any prior knowledge of the topic. Our discussion will be applicable only to $\RR^n$
 and 
will be, as said in the introduction, pragmatic and lacking  rigor. This will make the way some ideas are 
introduced somewhat awkward from a fully mathematical point of view. The interested reader is referred,
for example, to \cite{M} for a more consistent treatment.} in $\RR^n$.}


The symbol $\subseteq$ means \textbf{subset}, and $A \subseteq B$ reads
``$A$ is a subset of $B$. 
The case $A= B$ is \emph{not} excluded (every set is a subset
of itself). If we want to say that $A$ is a subset of $B$ but cannot equal $B$ itself,
we use the symbol $\subset$, so $A \subset B$. Sometimes we also say
that ``$A$ is contained in $B$." The symbol $\not \subset$ means ``not a subset,"
e.g. $\RR^3 \not \subset \RR^2$.

Recall that $(a,b) \subset \RR$ is an open interval, whereas $[a,b] \subset \RR$
is a closed interval. $(a,b]$ and $[b,a)$ are neither open nor closed.
Notice that an open interval has the following property. Pick any $x \in (a,b)$.
Then, we can always find another interval $I_x$ containing $x$ such that
$I_x \subset (a,b)$. Said in a slightly different way, given any $x \in (a,b)$,
we can always find a number $\ve > 0$ such that the interval $(x - \ve, x + \ve)$
is contained in $(a,b)$, i.e., $(x-\ve, x+\ve) \subset (a,b)$. A very similar idea
is used to talk about open sets in $\RR^n$. For this, we need
the following notation and terminology, which will be adopted from now on.
The \textbf{ball of radius $r$ and center $x_0$} in $\RR^n$ is defined as
\begin{gather}
B_r(x_0)= \Big \{ x \in \RR^n \, \Big | \, |x - x_0| < r \Big \}.
\label{ball_r}
\end{gather}
Sometimes we write $B^n_r(x_0)$ to emphasize that this is a subset of $\RR^n$,
and the case $x_0=0$ is sometimes abbreviated $B_r$ --- in which case
we refer simply to the ball of radius $r$.

A set $U \subset \RR^n$ is called \textbf{open} if for any $x \in U$ 
there exists a $r>0$ such that $B_r(x) \subset U$. Notice that $B_r(x)$ is always open
for any $x \in \RR^n$.
The \textbf{complement of a set}  $U \subset \RR^n$, denoted $U^c$,
 is the set of elements in 
$\RR^n$ that do \emph{not} belong to $U$. More precisely,
\begin{gather}
U^c = \Big \{ x \in \RR^n \, \Big | \, x \notin U \Big \}.
\nonumber
\end{gather}
Notice that this is the same as
\begin{gather}
U^c = \RR^n \backslash U.
\nonumber
\end{gather}
A set is called \textbf{closed} if its complement is open.
As an instructive exercise, the reader is invited to show that the set 
\begin{gather}
S^1= \Big \{ (x,y) \in \RR^2 \, \Big | \, x^2 + y^2 = 1 \Big \}
\nonumber
\end{gather}
is closed, by first identifying its complement, and then showing that it is open.

The \textbf{intersection} of two sets $A$ and $B$ contained in $\RR^n$, 
denoted $A \cap B$, is the set
of all elements that belong simultaneously to $A$ and $B$. We write,
\begin{gather}
A \cap B = \Big \{ x \in \RR^n \, \Big | \, x \in A, x \in B \Big \}.
\nonumber
\end{gather}
The reader can verify the following examples:
\begin{gather}
B_\frac{1}{2} \cap B_1 = B_\frac{1}{2},
\nonumber
\end{gather}
\begin{gather}
(-2,2) \cap (0,4) = (0,2),
\nonumber
\end{gather}
and
\begin{gather}
B_1 \cap B_\frac{1}{2}^c = \Big \{ (x,y) \in \RR^2 \, \Big | \, \frac{1}{4} \leq  x^2 + y^2 < 1 \Big \}.
\nonumber
\end{gather}
When $A$ and $B$ have no element in common, their intersection results in the \textbf{empty set},
denoted $\varnothing$. For example, let $x=(2,2,2) \in \RR^3$ and $y=(-2,-2,-2) \in \RR^3$, then
\begin{gather}
B_1(x) \cap B_1(y) = \varnothing.
\nonumber
\end{gather}
Inductively, we can define the intersection of more than two sets, in which case we write
$A \cap B \cap C$, etc. In fact, we can take the intersection of infinitely many sets.
For example, consider the sets $I_i \subset \RR$ defined by
\begin{gather}
I_i = [-\frac{1}{i}, \frac{1}{i} ],\, i=1,2,3,\dots
\nonumber
\end{gather}
Then
\begin{gather}
\bigcap_{i=1}^\infty I_i = \{0\}.
\nonumber
\end{gather}
To convince yourself of the above, draw the first few $I_i$'s and see what results from their 
intersection.

Given a set $U \subset \RR^n$, we can consider the ``smallest" closed set containing $U$.
To make this notion more precise, let $C(U)$ be the collection of all closed sets
containing $U$. I.e., $F \in C(U)$ if, and only if, $F$ is a closed set and $U \subset F$.
If we consider the intersection of all $F$'s with these properties, then we obtain a set 
called the \textbf{closure} of $U$, denoted $\overline{U}$. Thus,
\begin{gather}
\overline{U} = \bigcap_{F \in C(U)} F.
\nonumber
\end{gather}
To understand this notion, consider that, if $U$ is a closed set, then its closure is $U$
itself, i.e., $\overline{U} = U$ whenever $U$ is closed. If $U$ is not closed, then $\overline{U}$
consists of $U$ plus the points that are ``missing" to make $U$ closed. For instance, 
$[0,1)$ is not closed, because the endpoint $1$ is not included, thus
\begin{gather}
\overline{[0,1)} = [0,1].
\nonumber
\end{gather}
Different sets can have the same closure, e.g., the closure of $(0,1)$, $(0,1]$,
and $[0,1)$ are all equal to $[0,1]$. The reader should notice that
$\overline{U}$ is \emph{always} a closed set\footnote{It is an instructive exercise
to show this, as follows. By definition, one has to show that the complement
of $\overline{U}$ is open. This is done using the so-called de Morgan's law (see, 
for instance, \cite{H}): $\overline{U}^c = (\cap_F F)^c = \cup_F F^c$. Since
each $F^c$ is open because each $F$ is closed, and since (as the reader can check) the union
of open sets is open, we conclude that $\overline{U}^c$ is open, and thus
$\overline{U}$ is closed.}.

As another example, consider $B_1$, whose definition was given in 
(\ref{ball_r}) and in the text that immediately followed. $B_1$ is not closed because,
by its definition, it includes only the points whose distance to the origin 
is \emph{less} than one, i.e., it is ``missing" the points whose distance is exactly
\emph{equal} to one. Thus
\begin{gather}
\overline{B_1} = \Big \{ x \in \RR^n \, \Big | \, |x| \leq 1 \Big \}.
\nonumber
\end{gather}
Analogously, 
\begin{gather}
\overline{B_r(x_0)}= \Big \{ x \in \RR^n \, \Big | \, |x-x_0| \leq r \Big \}.
\nonumber
\end{gather}
Notice that the crucial difference of the above to (\ref{ball_r}) is that $<$ has
been replaced by $\leq$.

When forming $\overline{B_r(x_0)}$, the points that were ``missing" 
consists of a $n-1$-dimensional sphere of radius $r$ and center $x_0$ 
(if that is not clear, draw a picture of $\overline{B_r(x_0)}$ in two and three
dimensions). Therefore we define the \textbf{$n-1$ sphere of radius $r$ and center $x_0$}
 inside $\RR^n$ as
\begin{gather}
S_r(x_0) = \Big \{ x \in \RR^n \, \Big | \, |x-x_0| = r \Big \}.
\nonumber
\end{gather}
We write $S^{n-1}_r(x_0)$ if we want to emphasize that 
$S_r(x_0)$ is a subset\footnote{Notice that we write
$S^{n-1}_r(x_0)$, and not $S^{n}_r(x_0)$, because $S_r(x_0)$
always has  ``one less dimension" than $\RR^n$. For example, in two-dimensions, 
the ball of radius one is $\{ (x,y) \in \RR^2 \, \big | \, x^2 + y^2 \leq 1 \}$, but the sphere is only 
the circle $\{ (x,y) \in \RR^2 \, \big | \,  x^2 + y^2 = 1 \}$.
Although we shall not define the concept of dimension for general sets, this should
be intuitive. Also, see the comments at the end of this section\label{dimension}.}
 of $\RR^n$, and $S_r$ for $S_r(0)$ --- in this last case we refer simply
 to the sphere of radius $r$. The reader should check that 
 $S_r(x_0)$ is a closed set.
 
$S_r(x_0)$ can be thought of as the outermost points of $\overline{B_r(x_0)}$, or
the ``boundary" of $B_r(x_0)$. Furthermore, $\overline{B_r(x_0)}$ can be decomposed
into two pieces, one open and one closed, namely, $B_r(x_0)$ and $S_r(x_0)$, and such
that  $B_r(x_0) \cap S_r(x_0) = \varnothing$. Next, we define some similar notions
for sets that are not necessarily a ball. To do that, we need to first recall what
the union of sets is.

The \textbf{union} of of two sets $A$ and $B$ contained in $\RR^n$,
denoted $A \cup B$,  is defined
as the set of points that belong to $A$ or $B$, i.e.
\begin{gather}
A \cup B = \Big \{ x \in \RR^n \, \Big | \, x \in A, \, \text{or} \, x \in B \Big \}.
\nonumber
\end{gather}
For example, $(0,2) \cup (1,3) = (0,3)$, and $B_1 \cup S_1 = \overline{B_1}$.
As with intersections, we can consider the union of more than two sets, and even
of infinitely many sets.

Given a set  $U \subset \RR^n$, we can consider the ``largest" open set contained in $U$.
More precisely, let $O(U)$ be the collection of all open sets contained in $U$. I.e., 
$E \in O(U)$ if, and only if, $E$ is an open set and $E \subset U$.
The \textbf{interior} of $U$, denoted $\mathring{U}$, is defined as
\begin{gather}
\mathring{U} = \bigcup_{E \in O(U)} E.
\nonumber
\end{gather}
We point out that $\mathring{U}$ is \emph{always} an open set.

The \textbf{boundary} of a set $U \subset \RR^n$, denoted $\partial U$, is defined
by
\begin{gather}
\partial U = \overline{U} \backslash \mathring{U}.
\nonumber
\end{gather}
From the definition, it follows that $\partial U \cap \mathring{U} = \varnothing$.

Another important concept is that of connectedness.
A set $U \subset \RR^n $ is said to be \textbf{connected} 
if any two points in $U$ can be joined by a \emph{continuous} 
curve\footnote{To be precise, this is actually the definition of what we call
a \emph{path connected} set. Giving the precise definition of connectedness
would require a more extensive discussion. We point out, however, that in 
the main case of interest, i.e., when the set $U$ is open, the concepts of 
connectedness and 
path connectedness agree. The interested reader is referred to \cite{M} for details.}.
The basic intuition is that a connected set cannot be split into two parts that 
``do not communicate with each other." For instance, $B_r$ is a connected set, 
as are $(0,1) \subset \RR$, $[1,3)$, and $S_r$. The sets
$\RR \backslash \{0\}$, and $(-2,-1) \cup  (1,2)$, on the other hand, are
not connected.

We can now define one of the main types of sets of interest in this course. 
A \textbf{domain} in $\RR^n$ is a \emph{connected} and \emph{open} 
set\footnote{The concept of domain here has nothing to do with the domain of a function.}.
The notation $\Om$ will always denote a domain, unless stated otherwise.
The balls $B_r(x_0)$ are examples of domains. The spheres $S_r(x_0)$
are not domains because they fail to be open.

Given a domain $\Om$, we can consider its boundary $\partial \Om$ as defined earlier.
A domain is said to have \textbf{smooth boundary} if the following property holds:
\emph{locally}, $\partial \Om$ can be written as the graph of an infinitely differentiable
function. 

Probably the best way to understand this definition is via the following example. Consider the ball
of radius one in $\RR^3$. Recall that it is the set
\begin{gather}
B_1 = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, \sqrt{x^2 + y^2 + z^2 } < 1 \Big \}.
\nonumber
\end{gather}
Its boundary is the two-dimensional sphere
\begin{gather}
S_1 = \Big \{ ((x,y,z) \in \RR^3 \, \Big | \, \sqrt{x^2+  y^2 + z^2 } = 1 \Big \},
\nonumber
\end{gather}
i.e., $\partial B_1 = S_1$. Notice that $S_1$ cannot be the graph of a function, as it 
fails the analogue of the vertical line test in three-dimensions (any straight line 
through $S_1$ will cross it in two points). However, small pieces (this is roughly the meaning 
of ``locally") of $S_1$ can always be written
as a graph, as we now show.

Consider first the upper hemisphere of $S_1$, which we denote by $S_1^+$. It corresponds to the points that satisfy 
not only $\sqrt{x^2+  y^2 + z^2 } = 1$ but also that are ``above" the $xy$-plane, i.e.,
$z\geq 0$. Therefore, $S_1^+$ is given by the points that satisfy
\begin{gather}
 \sqrt{x^2+  y^2 + z^2 } = 1 \, \text{ and } \, z \geq 0.
\nonumber
\end{gather}
Solving for $z$ and using $z \geq 0$ to pick the positive square root gives
\begin{gather}
z = \sqrt{1 - x^2 - y^2}.
\nonumber
\end{gather}
But this defines $z$ as a function of $x$ and $y$. Therefore, the upper hemisphere
$S_1^+$ is the graph of the function
$h(x,y) = \sqrt{1 - x^2 - y^2}$, whose domain\footnote{For technical reasons, 
when one says that $\partial \Om$ is locally the graph of a smooth function, it is convenient
to define the domain of the function giving the graph to be always an open set. In this case
we would take $\{ (x,y) \in \RR^2 \, \big | \,x^2 + y^2 < 1 \}$. We shall, however, avoid this kind of subtlety here.}
 is the set $\{ (x,y) \in \RR^2 \, \big | \,
x^2 + y^2 \leq 1 \}$.

Similarly, the lower hemisphere, denoted by $S_1^-$, is given by
\begin{gather}
 \sqrt{x^2+  y^2 + z^2 } = 1 \, \text{ and } \, z \leq 0.
\nonumber
\end{gather}
Solving for $z$ and using $z \leq 0$ to pick now the negative square root gives
\begin{gather}
z = - \sqrt{1 - x^2 - y^2}.
\nonumber
\end{gather}
Again,  $z$ is a function of $x$ and $y$, whose domain
 is the set $\{ (x,y) \in \RR^2 \, \big | \,
x^2 + y^2 \leq 1 \}$. Since $S_1^+$ and $S_1^-$ completely cover the sphere, i.e.
$S_1^+ \cup S^1_- = S_1$, we have shown that locally $S^1$ can always be written as the graph of 
a function.

The above does not quite show yet that $B_1$ has smooth boundary. To do so, we need to show 
that the functions whose graphs give $S_1$ are infinitely differentiable. Consider the case
$S_1^+$. Taking derivatives of $\sqrt{1 - x^2 - y^2}$, we will find expressions that involve
\begin{gather}
\frac{1}{\sqrt{1 - x^2 - y^2}}.
\nonumber
\end{gather}
The above, and its derivatives, will be well-defined as long as 
$x^2 + y^2 < 1$, but for points such that $x^2 + y^2 = 1$, we would be dividing by zero.
The same statement holds for $S_1^-$.

Noticing that the points satisfying $x^2 + y^2 = 1$ correspond exactly to the equator
of the sphere, we conclude that the above shows that $S_1$ can always be written
locally as the graph of an infinitely differentiable function, \emph{except} for the 
points on the equator. To remedy this problem, we simply notice that there is nothing
in the definition of a smooth boundary that requires us to limit ourselves to only two 
pieces of $\partial \Om$. The points on the equator can also be shown to be part of
 graphs of certain infinitely differentiable functions, except that now $z$ will be one
 of the variables and $x$ or $y$ will be the function. For instance, 
 a similar reasoning as above shows that if we consider the right hemisphere, given by
\begin{gather}
 \sqrt{x^2+  y^2 + z^2 } = 1 \, \text{ and } \, x \geq 0,
\nonumber
\end{gather}
and the left hemisphere, given by
\begin{gather}
 \sqrt{x^2+  y^2 + z^2 } = 1 \, \text{ and } \, x \leq 0,
\nonumber
\end{gather}
 then we can solve for $x$ in terms of $y$ and $z$, obtaining the functions
\begin{gather}
x =  \sqrt{1 -   y^2 - z^2 }
\nonumber
\end{gather}
and 
\begin{gather}
x =  -\sqrt{1 -   y^2 - z^2 }.
\nonumber
\end{gather}
These functions both have  domain $y^2 + z^2 \leq 1$, and they are
 infinitely differentiable, except
 for the points satisfying $y^2 + z^2 = 1$. These points constitute the 
``principal meridian" of the sphere. But notice that these points are either on the
upper hemisphere or on the lower hemisphere, which we have already shown to be 
the graph of an infinitely differentiable function except for points on the equator.
However, the only points that are simultaneously on the equator and on the principal
meridian are the points $(1,0,0)$ and $(-1,0,0)$.
Therefore, we have shown that all points on the sphere, except those two,
belong to the graph of an infinitely differentiable function.
The reader can now probably imagine how those last two points are shown to also 
satisfy this property: one writes $y$ as a (two) function(s) of $x$ and $z$, and argue
analogously.

The conclusion is that by using enough ``caps" we can cover the sphere with pieces that are
always the graph of an infinitely differentiable function, as shown in figure 
\ref{caps}.

\begin{figure}[!ht]
\hskip 0.0cm \rotatebox{0.0}{\includegraphics[scale=.5]{Caps.pdf}}
\caption{Covering the sphere with several caps (credit: M. P. do Carmo, \emph{Riemannian
Geometry.} Birkh\"auser (1992)).}
\label{caps}
\end{figure}

 Hence, $B_1$ is a domain with smooth boundary.
Of course, there is nothing special about the radius equal to one or the origin as the center.
Nor is there anything special about the dimension $n=3$. Similar arguments can be used 
to show that $B_r(x_0) \subset \RR^n$ is always a domain with smooth boundary.

Some readers may find this last discussion too lengthy or complicated, but 
they should be able to grasp it quickly after working out some of the details by themselves.
We insist that this last example be understood thoroughly, as $B_1$ is the
prototypical example of domains with smooth boundary --- which will be one of the most
important sets used in the course.

A domain is called \textbf{bounded} if it can be enclosed inside a
ball of radius $r$ for some $r > 0$. For example,
the region
\begin{gather}
A = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, \frac{1}{16} < 
x^2 + y^2 + z^2 < 1 \Big \}
\nonumber
\end{gather}
is bounded because it lies inside\footnote{Of course, 
$A$ also lies inside the ball of radius $3$, or $1.5$... The definition 
only requires that it belongs to a ball of radius $r$ for \emph{some} $r> 0$.}
 the ball of radius $2$; while 
\begin{gather}
\widetilde{A} = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, x  > 0, y > 0, 
z > 0 \Big \}
\nonumber
\end{gather}
is not bounded.

Lastly, we make a comment about dimension. It should be intuitive that
$B_r \subset \RR^n$ has $n$ dimensions, whereas $S_r$ has $n-1$ dimensions
(the reader can consider the case $n=3$ for simplicity if necessary). This is because
in $B_r$ we have $n$ variables $x_1, x_2, \dots x_n$, but in $S_r$ one variable can
always be written as a function of the remaining $n-1$ variables, as we have shown above.
Hence, there are truly only $n-1$ independent coordinates in $S_r$. Although 
we shall not define the concept of dimension here, the reader should keep this intuitive
notion in mind: that a domain $\Om$ in $\RR^n$ is a $n$-dimensional spaces, whereas
its boundary $\partial \Om$ is a $n-1$-dimensional space\footnote{To be more precise,
the boundary is a space of \emph{at most} $n-1$ dimensions. However, in all
cases relevant for this course, it will in fact consist of a $n-1$ dimensional 
space.}. This will be important later
on when we discuss functions defined on $\Om$ and $\partial \Om$, i.e., 
$f: \Om \rar \RR$ and $g: \partial \Om \rar \RR$, respectively.
$f$ is then a function of $n$ variables, while $g$ is a function of $n-1$ varaibles.

\section{Partial derivatives.}
In this section we recall some basic notions about partial derivatives.
The reader can check \cite{S} for a more detailed review of the topics here
presented.
The \textbf{partial derivative} with respect to the $i^\text{ih}$ coordinate will be denoted by
\begin{gather}
\frac{\partial}{\partial x_i}, \, \text{ or } \, \partial_i,
\nonumber
\end{gather}
with higher order derivatives denoted accordingly, e.g., 
\begin{gather}
\frac{\partial^2}{\partial x_i \partial x_j}, \, \partial^2_{ij},
\, \text{ or simply } \, \partial_{ij}.
\nonumber
\end{gather}
We sometimes speak simply of ``derivative" to mean ``partial derivative."
If $f$ is a function of $n$ variables (for instance, a function defined in $\RR^n$), i.e.,
$f(x_1, x_2, \dots, x_n)$, its derivative with respect to the $i^\text{ih}$ coordinate
is also denoted by a subscript:
\begin{gather}
f_i = \partial_i f,
\nonumber
\end{gather}
or yet
\begin{gather}
f_{x_i} = \partial_i f.
\nonumber
\end{gather}
Notice that if $x \in \RR^n$, then $f(x)$ means that $f$ is a function of $n$
variables, i.e., $f(x) = f(x_1, x_2, \dots, x_n)$, in which case
$f_i(x)$ means the derivative of $f$ with respect to the $i^\text{ih}$ coordinate
evaluated
at $x$. You should be careful not to confuse the notation  $f_i$
for partial derivatives with the notation for the $i^\text{th}$ component of
a vector valued function.

A function is said to be \textbf{$k$-times differentiable} if all its partial derivatives up
to order $k$ exist, and \textbf{$k$-times continuously differentiable} if all
its partial derivatives up to order $k$ exist and are continuous.
It is important to notice the difference between these two  concepts.
The reader is encouraged to try to find an example of a function  
whose derivative exists but is not continuous; i.e., find $f: \RR \rar \RR$,
such that $f^\prime$ is well-defined, but $f^\prime$ is not a continuous function.
We denote by $C^k(\Om)$ the set of real-valued $k$-times continuously differentiable
functions defined on $\Om$ (recall that $\Om$ is always a domain in $\RR^n$). 
More precisely,
\begin{gather}
C^k(\Om) = \Big \{ f: \Om \rar \RR \,\, \Big | \, \text{ all partial derivatives of } f \text{ up to order } k \text{ exist and are continuous }
 \Big \}.
 \nonumber
\end{gather}
The sets $C^k(\overline{\Om})$ and $C^k(\partial \Om)$ are defined 
similarly\footnote{The reader should make sure that he or she understands 
the definition of derivative
of a function defined on a \emph{closed} set, such as $\overline{\Om}$.}.

From now on, it will be assumed that the functions involved are sufficiently
differentiable, so that all the formulas involving derivatives will make sense.

A very important tool to compute derivatives is provided by the 
\textbf{chain rule}. Recall that if $f: \RR \rar \RR$ and
$g: \RR \rar \RR$ are differentiable, then
\begin{gather}
(f\circ g)^\prime (x) = f^\prime(g(x))g^\prime(x),
\nonumber
\end{gather}
where $f\circ g$ is the \textbf{composition} of $f$ and $g$. We can omit $x$ and write the above 
as
\begin{gather}
(f\circ g)^\prime  = (f^\prime \circ g) g^\prime.
\nonumber
\end{gather}
Next, we recall how this rule generalizes to functions of several variables.

Consider $g: (a,b) \rar \Om$ and $f: \Om \rar \RR$, so that $f \circ g: \RR \rar \RR$ 
is well-defined. We can write $g=(g_1,g_2, \dots, g_n)$. Thus,
\begin{gather}
(f\circ g)^\prime(x) = \sum_{i=1}^n \partial_i f(g(x)) g_i^\prime(x),
\nonumber
\end{gather}
or simply
\begin{gather}
(f\circ g)^\prime = \sum_{i=1}^n (\partial_i f \circ g) g_i^\prime.
\nonumber
\end{gather}
Sometimes one sees the above written  as 
\begin{gather}
(f\circ g)^\prime = \sum_{i=1}^n \partial_i f  g_i^\prime.
\nonumber
\end{gather}
In this last expression, it is implicitly understood that for each $x \in (a,b)$,
$\partial_i f$ is to be evaluated at $g(x)$, i.e., $\partial_if(g(x))$, even though
the composition $\circ g$ in $\partial_i f \circ g$ has been omitted. Although
this may be a bit confusing at first sight, it is the only thing that makes sense,
since we are computing the derivative of the composition $f \circ g$. 

There is a simple mnemonics for the chain rule. Recall that for single variable
functions, we can write $x=g(t)$ and $f = f(x)$, so that
\begin{gather}
\frac{d}{dt}(f\circ g) = \frac{df}{dx}\frac{dx}{dt},
\label{chain_1}
\end{gather}
where of course $\frac{dx}{dt} = g^\prime$.

In the case $g: (a,b) \rar \Om$ and $f: \Om \rar \RR$, 
let us write $x=g(t)$, so that $x=(x_1,x_2, \dots, x_n) =
(g_1, g_2, \dots, g_n)$, i.e., $x_i(t) = g_i(t)$. Then
\begin{gather}
(f\circ g)^\prime = \sum_{i=1}^n \frac{\partial f}{\partial x_i} 
\frac{d x_i}{d t},
\label{chain_n}
\end{gather}
where $\frac{d x_i}{d t} = g_i^\prime$. Comparing (\ref{chain_1})
with (\ref{chain_n}), it is seen that they have exactly the same form,
except that because $f$ is a function of $n$ variables, in (\ref{chain_n})
the derivatives of $f$ are partial derivatives and we have to sum
over the different components $x_i$. Notice that each $x_i = g_i$ is a function
of only one variable (the variable $t$). Notice also that with this notation, 
there is no need to write the composition $\circ g$ along with the partial derivative
of $f$: since  the coordinates in $\Om$ are denoted by $x$, and we wrote
$\frac{\partial f}{\partial x_i}$, it is implicitly understood that this is evaluated
at $x \in \Om$; but $x = g(t)$.

Sometimes, we shall also need to compute the derivative of the composition of
two functions of several variables. Suppose $\Om_0 \subset \RR^m$ and $\Om_1 \subset \RR^n$
are two domains in $\RR^m$ and $\RR^n$, respectively (notice that $m$ and $n$ can be different).
Let $g : \Om_0 \rar \Om_1$ and $f: \Om_1 \rar \RR$, so that $f \circ g$ is well-defined.
Notice that $f \circ g$ is a function of $m$ variables, thus, when computing
derivatives of $f\circ g$ we have to talk about its \emph{partial} derivatives.
To do so, we apply formula (\ref{chain_n}) \emph{for each partial derivative}.
More precisely, denote the coordinates in $\Om_0$ by $x$, so 
$x = (x_1,x_2,\dots,x_m) \in \Om_0$, and the coordinates
in $\Om_1$ by $y$, so $y=(y_1,y_2,\dots, y_n) \in \Om_1$.
Then we have $y=g(x)$ for each $x \in \Om_0$, i.e., $y_i = g_i(x)$, $i=1,2, \dots, m$.
The $i^{th}$ partial derivative
of  $f \circ g$ is given by
\begin{gather}
\frac{\partial (f \circ g)}{\partial x_i} 
= \sum_{j=1}^n \frac{\partial f}{\partial y_j} \frac{\partial y_j}{\partial x_i},
\label{chain_n_m}
\end{gather}
where $\frac{\partial y_j}{\partial x_i} = \frac{\partial g_j}{\partial x_i}$.
Notice that again, by the conventions we adopted to indicate the coordinates
in $\Om_0$ and $\Om_1$, there is no need to write $\circ g$ along with 
$\frac{\partial f}{\partial y_j} $ since $y_j = g_j(x)$.
The reader should compare (\ref{chain_n_m}) with (\ref{chain_n}) and realize
that the latter is a particular case of the former when $m=1$.

If $f:\Om \subset \RR^n \rar \RR$, its \textbf{gradient}, denoted $\nabla f$,
 is the
$n$-component vector defined as
\begin{gather}
\nabla f = (\partial_1 f, \partial_2 f, \dots, \partial_n f).
\nonumber
\end{gather}
And if $f: \Om \subset \RR^n \rar \RR^m$, its \textbf{Jacobian matrix},
denoted $Df$, is given by
\begin{gather}
Df =
\begin{bmatrix}
 \partial_1 f_1 & \partial_2 f_1 & \cdots & \partial_n f_1 \\
 \partial_1 f_2 & \partial_2 f_2 & \cdots & \partial_n f_2 \\
 \vdots & & & \vdots \\
 \partial_1 f_m & \partial_2 f_m & \cdots & \partial_n f_m 
\end{bmatrix}
\nonumber
\end{gather}
Other notations for $Df$ are
\begin{gather}
\frac{\partial (f_1,f_2,\dots, f_m)}{\partial (x_1, x_2, \dots, x_n)}
\, \text{ or }  \, (\partial_i f_j).
\nonumber
\end{gather}
With these notations, it is instructive to show that (\ref{chain_n}) is given
by
\begin{gather}
(f \circ g)^\prime = \langle \nabla f \circ g, g^\prime \rangle,
\nonumber
\end{gather}
and that (\ref{chain_n_m}) can be written as
\begin{gather}
D(f \circ g) = \nabla f \circ g \cdot Dg.
\nonumber
\end{gather}
In this last expression, $\cdot$ is simply the matrix multiplication of the 
$1 \times n$ matrix $\nabla f \circ g$ by the $n \times m$ matrix $Dg$. 
The result is a $1 \times m$ matrix whose $i^{th}$ column is the $i^{th}$
partial derivative of $f\circ g$. The attentive reader has probably already
noticed that $\nabla f$ equals $D f$ when $f: \RR \rar \RR^n$.
Therefore, \emph{all of the above formulas for the chain rule are particular cases of the general
formula:}
\begin{gather}
D(f \circ g) = D f \circ g \cdot Dg.
\nonumber
\end{gather}

\section{Integrals.}
Here we recall some basic facts about integrals of functions
of several variables. It will be assumed that the functions satisfy all
the required hypotheses to make the integrals involved well-defined.
The reader can check \cite{S} for a more detailed review of the topics here
presented.

The \textbf{integral} of a function $f: \Om \subset \RR^n \rar \RR$
will be denoted by
\begin{gather}
\int_\Om f.
\nonumber
\end{gather}
Notice that we do not write the volume element $dV$ or $d\vec{x}$
for this multidimensional integral, although sometimes it may be convenient
to do so (e.g., to stress which ones are the variables of integration), in which
case we shall write
\begin{gather}
\int_\Om f (x) \, dx.
\nonumber
\end{gather}
Notice, also, that $dx$ represents the volume element in $n$-dimensions, i.e., 
\begin{gather}
dx = dx_1 dx_2 \cdots dx_n,
\nonumber
\end{gather}
so $\int_\Om f (x) \, dx$ can be written more explicitly as
\begin{gather}
\int_\Om f(x_1, x_2, \dots, x_n) \, dx_1  dx_2 \cdots dx_n.
\nonumber
\end{gather}
The above also implies that we avoid
 using the notations of ``several integrals."
For example, if $\Om$ is a domain in $\RR^2$, we do \emph{not} write,
\begin{gather}
 \iint\limits_\Om f(x_1,x_2) \, dx_1 dx_2
\nonumber
\end{gather}
since, once it is known that $\Om \subset \RR^2$, it is superfluous 
to write the integral sign twice.

Consider a domain $\Om$  with smooth boundary $\partial \Om$. 
Remember that it is possible
to carry out integration of functions defined over $\partial \Om$, which we write,
\begin{gather}
\int_{\partial \Om} g,
\nonumber
\end{gather}
where $g: \partial \Om \rar \RR$. As before, we do not use
some of the ``standard" notation found in calculus books, avoiding
writing $dA$, $d\vec{A}$, etc. for the area element of $\partial \Om$. When
it is necessary to write such an area element, we denote it by $ds$, 
\begin{gather}
\int_{\partial \Om} g \, ds,
\nonumber
\end{gather}
or 
\begin{gather}
\int_{\partial \Om} g(x) \, ds.
\nonumber
\end{gather}
Intuitively, $ds$ is the ``restriction" of the $n$-dimensional
volume element $dx$ to the $n-1$-dimensional boundary
$\partial \Om$. For instance, if $\Om \subset \RR^3$
is the upper-half plane, 
\begin{gather}
\Om = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, z > 0 \Big \},
\nonumber
\end{gather}
then $\partial \Om$ is the $xy$-plane,
\begin{gather}
\partial \Om = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, z = 0 \Big \},
\nonumber
\end{gather}
and in this case $ds = dx dy$. It is important to remark that
even if the boundary is $n-1$-dimensional, 
$ds$ is still referred to as an \emph{area} element --- although sometimes we 
also use \emph{volume element induced on the boundary}, \emph{induced 
volume element}, or \emph{boundary volume element}.

Again, the reader should notice that we avoid some of the more involved notation
for the boundary integrals. For instance, if $\Om$ is two-dimensional,
the $\partial \Om$ is a curve, and $\int_{\partial \Om}$ is sometimes written
$\oint_{\partial \Om}$. This notation will \emph{not} be employed here.

In some exceptional situations, one wants to integrate
a function of $n+m$ variables with respect to, say, the first $n$ variables.
In such cases we write,
\begin{gather}
f = f(x,y) = f(x_1, x_2, \dots, x_n, y_1, y_2, \dots y_m ),
\nonumber
\end{gather}
and the integrals
\begin{gather}
\int_\Om f(x,y) \, dx,
\nonumber
\end{gather}
and
\begin{gather}
\int_{\partial \Om} g(x,y) \, ds(x),
\nonumber
\end{gather}
where $\Om \subset \RR^n$. I.e., even though the functions
$f$ and $g$ involve $n+m$ variables, we are considering an integral over
a domain $\Om$ that belongs to $\RR^n$ --- thus, we have integrals 
over the first $n$-variables. In the above, the $x$ in $ds(x)$ is used
to emphasize that  only the first $n$ variables, encoded in $x$,
enter in the integration.

Consider the domain $\Om \subset \RR^3$ given by
\begin{gather}
\Om = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, z \geq 0 \Big \}.
\nonumber
\end{gather}
Then, as in the previous example, $\partial \Om$ is the $xy$-plane,
\begin{gather}
\partial \Om = \Big \{ (x,y,z) \in \RR^3 \, \Big | \, z = 0 \Big \}.
\nonumber
\end{gather}
Given a function $f: \Om \rar \RR$, one naturally
gets a function defined on $\partial \Om$ by simply setting $z=0$, 
i.e., the function $f(x,y,0)$ is a function defined on the boundary
$\partial \Om$. The same idea works for a general domain, as we next explain.

A function $f: \overline{\Om} \rar \RR$ naturally defines a function 
on $\partial \Om$, called the \textbf{restriction} of $f$ to the boundary, 
denoted by $\left. f \right|_{\partial \Om}$, and given by
\begin{gather}
\left. f \right|_{\partial \Om}(x) = f(x),  \text{ for } x \in \partial \Om.
\nonumber
\end{gather}
From this it follows that we can also
integrate a function defined on $\overline{\Om}$ over the boundary , i.e., 
\begin{gather}
\int_{\partial \Om} f 
\nonumber
\end{gather}
is well-defined.

Another notion that needs to be recalled is that of 
the \emph{normal derivative}.
Given a domain $\partial \Om$ with smooth boundary and
$x \in \partial \Om$, the \textbf{normal vector}
to $\partial \Om$ at $x$, denoted\footnote{Calculus books usually
denote the normal vector by $N$, $n$, $\vec{N}$, or $\vec{n}$.} $\nu(x)$, or $\nu_x$ or yet
simply $\nu$, is defined as the vector based at $x$ that has unit length,
 is perpendicular to the tangent plane to $\partial \Om$ at $x$, 
and points towards the ``outside" of $\Om$.

For example, if $\Om$ is the domain 
\begin{gather}
\Om = \Big \{ (x,y) \in \RR^2 \, \Big | \, y > 0 \Big \}, 
\nonumber
\end{gather}
then 
\begin{gather}
\partial \Om = \Big \{ (x,y) \in \RR^2 \, \Big | \, y = 0 \Big \}, 
\nonumber
\end{gather}
(i.e., $\partial \Om$ is simply the $x$-axis) and for any
$(x,0) \in \partial \Om$, the normal is given
by $\nu = (0,-1)$.

The normal to the $n-1$-dimensional sphere $S_r$ at $x$ is
given by $\nu = \frac{1}{r} x$. The reader should also check that the normal
to $S_r(x_0)$ at $x$ is given by $\nu = \frac{1}{r}(x-x_0)$.

The \textbf{normal derivative} of a function $f : \overline{\Om} \rar \RR$
is a function on $\partial \Om$, denoted $\frac{\partial f}{\partial \nu}$
or $\partial_\nu f$, and defined by
\begin{gather}
\frac{\partial f}{\partial \nu}= \langle \nabla f , \nu \rangle,
\text{ on }  \partial \Om,
\nonumber
\end{gather}
or more explicitly,
\begin{gather}
\frac{\partial f}{\partial \nu}(x) = \langle \nabla f(x), \nu(x) \rangle,
\, x \in \partial \Om.
\nonumber
\end{gather}
The \textbf{Laplacian} of a function $f:\Om \subset \RR^n \rar \RR$, denoted $\Delta f$, 
is defined as
\begin{gather}
\Delta f = \frac{\partial^2 f }{\partial x_1^2}
+
\frac{\partial^2 f }{\partial x_2^2}
+ \cdots
+
\frac{\partial^2 f }{\partial x_n^2}.
\nonumber
\end{gather}
The above can be written in several equivalent ways, e.g.,
\begin{align}
\begin{split}
\Delta f &= \frac{\partial^2 f }{\partial x_1^2}
+
\frac{\partial^2 f }{\partial x_2^2}
+ \cdots
+ \frac{\partial^2 f }{\partial x_n^2} \\
& = \sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2} \\
& = \partial^2_{11} f + \partial^2_{22} f + \cdots + \partial^2_{nn} f \\
& = \sum_{i=1}^n \partial^2_{ii} f \\
& = f_{11} + f_{22} + \cdots + f_{nn} \\
& = \sum_{i=1}^n f_{ii}\\
& = f_{x_1 x_1} + f_{x_2 x_2} + \cdots + f_{x_n x_n} \\
& = \sum_{i=1}^n f_{x_i x_i}.
\end{split}
\nonumber
\end{align}

With the above definitions at hand, we can now recall
the following \textbf{Green's identities:}
\begin{gather}
\int_\Om \langle \nabla f, \nabla g \rangle 
= - \int_\Om f \Delta g + \int_{\partial \Om} f \frac{\partial g}{\partial \nu},
\nonumber
\end{gather}
and 
\begin{gather}
\int_\Om \left( g \Delta f - f \Delta g \right )  
=  \int_\Om \left(  g \frac{\partial f}{\partial \nu} - 
f \frac{\partial g}{\partial \nu} \right),
\nonumber
\end{gather}
where $f,g:  \overline{\Om}  \subset \RR^n\rar \RR$.
These two formulas can be derived from the formula
for \textbf{integration by parts} in $n$-variables:
\begin{gather}
\int_\Om g \partial_i f = - \int_\Om f \partial_i g + \int_{\partial \Om}
g f \nu_i,
\nonumber
\end{gather}
where $\nu_i$ is the $i^\text{th}$ component of the normal vector
$\nu = (\nu_1, \nu_2, \dots, \nu_n)$.

\section{Quantifiers and the formation of mathematical sentences.}
Here we introduce some mathematical symbols that are quite useful
to make shorthand notation. We also give some examples of their use, while
making some general remarks about the form of certain mathematical statements.
Once more, we emphasize that our discussion is informal and lacks
proper mathematical rigor\footnote{The reader interested in
a thorough discussion can consult  \cite{H,T}.}.

Below is a list of mathematical symbols that are often used, along with their
interpretation:
\begin{align}
\begin{split}
\text{\underline{Symbol}} & \hspace{1cm} \text{\underline{Reads as}} \\
\forall \hspace{1cm}  & \hspace{0.9cm} \text{ for all}\\
\exists \hspace{1cm}  & \hspace{0.9cm} \text{ there exists}\\
\Rightarrow \hspace{0.8cm} & \hspace{0.9cm} \text{ if... then} \\
\Leftrightarrow\hspace{0.8cm}  &  \hspace{0.9cm} \text{ if and only if}
\end{split}
\nonumber
\end{align}
Let us see some examples of how such symbols are employed, and also 
of how to structure mathematical statements using them.

For example,
\begin{gather}
x >  1 \Rightarrow x > 0
\nonumber
\end{gather}
reads ``if $x$ is greater than one, then it is greater than zero." While this
 statement is true (a number that is greater than one is also greater than 
 zero), the correct use of $\Rightarrow$ has nothing to do with 
whether the sentence is in fact true. In other words, the sentence
$x >  1 \Rightarrow x > 0$ draws a conclusion about $x$, namely,
that $x>0$, under the assumption that $x >1$. But whether
or not the $x$ in question is in fact greater than one is completely open.
Thus, 
\begin{gather}
x > 0  \Rightarrow x > 1
\nonumber
\end{gather}
is also a correct use of the symbol $\Rightarrow$, except that now the sentence
is false: from the knowledge that $x>0$, it cannot be concluded that 
$x>1$, since there are numbers that are greater than zero but are not
greater than one. Summing up, $\Rightarrow$ is used in form
\begin{gather}
\text{Claim 1} \Rightarrow \text{Claim 2}
\nonumber
\end{gather}
to state that \emph{if} Claim 1 is true, \emph{then} Claim 2 must also be true.
Whether or not Claim 1 is in fact true is not addressed. As a colloquial 
analogy, the reader can imagine a sentence like ``If it rains, 
then the floor gets wet." It does not say anything about the actual status
of the weather, i.e., whether it is raining or not. 

Also, the full statement ``if .... then ..." does not have to be true in order
to carry a correct use of $\Rightarrow$. Although we shall not define
what is meant by  ``correct use," the idea is that it is employed in a
sentence that makes sense. The statement 
$x > 0  \Rightarrow x > 1$ makes sense, i.e., we can read and understand it, 
even though it is a wrong statement (in fact, had we been unable to understand
what it says, we would not even be capable of saying whether it is right or wrong).
As another colloquial analogy, consider the statement
``If I drink clean water, then I will get very sick." This is a correct use of 
``if" and ``then": the sentence is well-formed, it makes sense, and it is 
grammatically correct --- despite the fact that its content is false.

One important thing to keep in mind is that even if 
$\text{Claim 1} \Rightarrow \text{Claim 2}$ is true, knowing that 
Claim 2 is true does not say anything about Claim 1. For instance,
consider the statement:
 \emph{if} a (differentiable) 
function $f$ has a local maximum at
$a$, then $f^\prime(a) =0$. This statement is true, as you learned in calculus.
However, knowing  that $f^\prime(a) =0$ does not give information 
about the nature of the point $a$; it could be a local maximum, a local minimum,
or neither (e.g., the derivative of $x^3$ at zero is zero, but zero is neither
a local maximum nor a local minimum). Using once more an analogy,
even if the statement ``If it rains, the floor gets wet" is true, 
we cannot conclude that it had rained by  noticing
that the floor is wet.

$\exists$ is used to indicate that there is at least one element in a set satisfying
a certain property. For instance,
\begin{gather}
 \exists x \in \RR, \, x^2 - 1 = 0
 \nonumber
 \end{gather}
 reads, ``there exists an element $x$ in the set of real numbers
 such that $x^2-1=0$." Notice that $\exists$ does not say that 
 ``there exists only one."  For instance, the above statement is true because
 $x=1$ satisfies $ x^2 - 1 = 0$, but $x=-1$ also satisfies the equation.
 Had we claimed that there existed only one element satisfying 
 $x^2 - 1 = 0$, then the statement would have been false.
 
As it happened for $\Rightarrow$, the right use of $\exists$ has nothing to
do with whether the statement is correct or not. For instance, 
\begin{gather}
 \exists x \in \RR, \, x^2 +1 = 0
 \nonumber
 \end{gather}
is false, since there is no real number $x$ satisfying the equation 
$ x^2 +1 = 0$. However, this was a perfect legitimate use of $\exists$, as the
statement $ \exists x \in \RR, \, x^2 +1 = 0$, while false, is something that
makes sense, in the sense that we can perfectly understand what is being 
claimed.

$\forall$ is used to indicate that certain statement being made is to be
applied for all elements under consideration. For instance,
\begin{gather}
\forall x \in (-1,\infty), \, x + 1 > 0
\nonumber
\end{gather}
reads ``for all $x$ belonging to $(-1,\infty)$, $x+1$ is greater than zero." 
We sometimes find more convenient to state this in the reverse order\footnote{Which,
the reader should notice, does not change the meaning of what is being said.}, i.e., 
\begin{gather}
 x + 1 > 0 \, \, \forall x \in (-1,\infty) ,
\nonumber
\end{gather}
reading ``$x+1$ is greater than zero for all $x$ belonging to 
$(-1,\infty)$." We use ``for any" as a synonym of $\forall$, so we could also have
said ``$x+1$ is greater than zero for any $x$ belonging to 
$(-1,\infty)$."  As in the previous examples, notice that the correct use of $\forall$
has nothing to do with whether the statement being made is true or false.

Finally, $\Leftrightarrow$ is used as follows:
\begin{gather}
\text{Claim 1} \Leftrightarrow \text{Claim 2}
\nonumber
\end{gather}
means that Claim 1 $\Rightarrow$ Claim 2 \emph{and}, reciprocally,
Claim 2 $\Rightarrow$ Claim 1. For instance, 
\begin{gather}
x^2 = 0 \Leftrightarrow x = 0
\nonumber
\end{gather}
reads ``$x^2$ is equal to zero if and only if $x$ is equal to zero," and it encodes
two statements: that if $x^2=0$ then $x=0$, and also that
if $x=0$ then $x^2=0$. Once more, we remark that while
the above statement is true, the correct use of $\Leftrightarrow$
is independent of this.


\begin{thebibliography}{ZZZZ}
\bibitem[H]{H} Halmos, P. R., \emph{Naive Set Theory.} Martino Fine Books (2011).
\bibitem[M]{M} Munkres, J., \emph{Topology.} Pearson, 2 edition (2000).
\bibitem[PR]{PR} Pinchover, Y.; Rubinstein, J., \emph{An 
introduction to partial differential equations.} Cambridge University Press (2005).
\bibitem[S]{S} Stewart, J., \emph{Calculus.} Cengage Learning; 7 edition (2012).
\bibitem[T]{T} Tarski, A.,  \emph{Introduction to Logic: and to the Methodology of Deductive Sciences.} Dover Publications (1995).
\end{thebibliography}

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