Finished moving files into Chapter 2

Also made some minor changes

IndexingSets.tex

@@ -0,0 +1,175 @@
+\begin{section}{Indexing Sets}
+
+Suppose we wish to consider the following collection of open intervals:
+\[
+(0,1), (0,1/2), (0,1/4), \ldots, (0,1/2^{n-1}), \ldots
+\]
+This collection has a natural way for us to ``index" the sets:
+\[
+I_1=(0,1), I_2=(0,1/2), \ldots, I_n=(0,1/2^{n-1}), \ldots
+\]
+In this case the sets are \textbf{indexed} by the set $\mathbb{N}$.  The subscripts on the capital letters are taken from the \textbf{index set}.  If we wanted to talk about an arbitrary set from this indexed collection, we could use the notation $I_n$.
+
+Let's consider another example:
+\[
+\{a\}, \{a,b\}, \{a,b,c\}, \ldots, \{a,b,c,\ldots,z\}
+\]
+An obvious way to index these sets is as follows:
+\[
+A_1=\{a\}, A_2=\{a,b\}, A_3=\{a,b,c\}, \ldots, A_{26}=\{a,b,c,\ldots,z\}
+\]
+In this case, the collection of sets is indexed by $\{1,2,\ldots, 26\}$.
+
+\begin{remark}
+Using indexing sets in mathematics is an extremely useful notational tool, but it is important to keep straight the difference between the sets that are being indexed, the elements in each set being indexed, the indexing set, and the elements of the indexing set.
+\end{remark}
+
+Any set (finite or infinite) can be used as an indexing set.  Often capital Greek letters are used to denote arbitrary indexing sets and small Greek letters to represent elements of these sets.  For example, we might use $\Delta$ (capital delta) to refer to an indexing set and write $\alpha \in \Delta$ for an individual index.  Typically, if the indexing set is some subset of $\mathbb{Z}$ (like $\mathbb{N}$), then we would use letters like $k,m,n,l$ for an individual index.  Likewise, if the indexing set is $\mathbb{R}$, then we might use $s,t,x,y$ as indices.  
+
+\begin{example}
+Here are some examples of common notation that you will encounter.
+\begin{enumerate}
+\item If $\Delta$ is a set and we have a collection of sets indexed by $\Delta$, then we may write
+\[
+\{S_{\alpha}\}_{\alpha\in \Delta}
+\]
+to refer to this collection.  We read this as ``the set of $S$-alphas over alpha in Delta."
+\item If a collection of sets is indexed by the natural numbers, then we may write
+\[
+\{U_n\}_{n\in\mathbb{N}}
+\]
+or
+\[
+\{U_n\}_{n=1}^{\infty}.
+\]
+\item Borrowing from this idea, we can write the collection $\{A_1,\ldots,A_{26}\}$ from the beginning of the section as
+\[
+\{A_n\}_{n=1}^{26}.
+\]
+
+\end{enumerate}
+\end{example}
+
+\begin{definition}
+Suppose we have a collection $\{A_{\alpha}\}_{\alpha\in\Delta}$.
+
+\begin{enumerate}
+\item The \textbf{union of the entire collection} is defined via
+\[
+\bigcup_{\alpha\in\Delta} A_{\alpha}=\{x:x\in A_{\alpha} \mbox{ for some }\alpha\in \Delta\}.
+\]
+
+\item The \textbf{intersection of the entire collection} is defined via
+\[
+\bigcap_{\alpha\in\Delta} A_{\alpha}=\{x:x\in A_{\alpha} \mbox{ for all }\alpha\in \Delta\}.
+\]
+
+\end{enumerate}
+\end{definition}
+
+\begin{example} 
+In the special case that $\Delta=\mathbb{N}$, we write
+\[
+\bigcup_{n=1}^{\infty}A_n= \{ x : x \in A_n \mbox{ for some } n \in \mathbb{N}\}= A_1\cup A_2 \cup A_3 \cup \cdots
+\] 
+and
+\[
+\bigcap_{n=1}^{\infty}A_n= \{ x : x \in A_n \mbox{ for all } n \in \mathbb{N}\} = A_1\cap A_2 \cap A_3 \cap \cdots
+\] 
+Similarly, if $\Delta=\{1,2,3,4\}$, then
+\[
+\bigcup_{n=1}^{4}A_n= A_1\cup A_2 \cup A_3 \cup A_4
+\] 
+and
+\[
+\bigcap_{n=1}^{4}A_n= A_1\cap A_2 \cap A_3 \cap A_4.
+\] 
+\end{example}
+
+\begin{remark}
+Notice the difference between ``$\bigcup$" and ``$\cup$" (respectively, ``$\bigcap$" and ``$\cap$").  The larger versions of the union and intersection symbols very much like the notation that you've likely seen for sums (e.g., $\displaystyle \sum_{i=1}^\infty i^2$).
+\end{remark}
+
+\begin{exercise}
+Let $\{I_n\}_{n\in\mathbb{N}}$ be the collection of open intervals from the beginning of the section.  Find each of the following.
+\begin{enumerate}
+\item $\displaystyle \bigcup_{n\in\mathbb{N}}I_n$
+\item $\displaystyle \bigcap_{n\in\mathbb{N}}I_n$
+\end{enumerate}
+\end{exercise}
+
+\begin{exercise}
+Repeat the previous exercise, but assume that the sets are closed intervals.
+\end{exercise}
+
+\begin{exercise}
+Let $\{A_n\}_{n=1}^{26}$ be the collection from earlier in the section.  Find each of the following.
+\begin{enumerate}
+\item $\displaystyle \bigcup_{n=1}^{26}A_n$
+\item $\displaystyle \bigcap_{n=1}^{26}A_n$
+\end{enumerate}
+\end{exercise}
+
+\begin{exercise}
+Let $S_n = \{x \in \mathbb{R} \ : \ n-1<x<n \}$, where $n\in \mathbb{N}$.  Find each of the following.
+\begin{enumerate}
+\item $\displaystyle \bigcup_{n=1}^{\infty}S_n$
+
+\item $\displaystyle \bigcap_{n=1}^{\infty}S_n$
+\end{enumerate}
+\end{exercise}
+
+\begin{exercise}
+Let $T_n = \{x \in \mathbb{R} \ : \ -\frac{1}{n}<x< \frac{1}{n} \}$, where $n\in \mathbb{N}$.  Find each of the following.
+\begin{enumerate}
+\item $\displaystyle \bigcup_{n=1}^{\infty}T_n$
+
+\item $\displaystyle \bigcap_{n=1}^{\infty}T_n$
+\end{enumerate}
+
+\end{exercise}
+
+\begin{exercise}
+For each $r\in\mathbb{Q}$ (the rational numbers), let $N_r$ be the set containing all real numbers \emph{except} $r$.  Find each of the following.
+\begin{enumerate}
+\item $\displaystyle \bigcup_{r\in\mathbb{Q}}N_r$
+
+\item $\displaystyle \bigcap_{r\in\mathbb{Q}}N_r$
+\end{enumerate}
+
+\end{exercise}
+
+\begin{definition}
+We say that a collection of sets $\{A_{\alpha}\}_{\alpha\in\Delta}$ is \textbf{pairwise disjoint} if $A_{\alpha} \cap A_{\beta}=\emptyset$ whenever $\alpha\neq \beta$.
+\end{definition}
+
+\begin{exercise}
+Draw a Venn diagram of a collection of 3 sets that are pairwise disjoint.
+\end{exercise}
+
+\begin{exercise}
+Provide an example of a collection of three sets, say $\{A_1, A_2, A_3\}$, such that the collection is \emph{not} pairwise disjoint, but 
+\[
+\bigcap_{n=1}^3 A_n=\emptyset.
+\]
+\end{exercise}
+
+\begin{theorem}[Generalized Distribution of Union and Intersection]
+Suppose we have a collection $\{A_{\alpha}\}_{\alpha\in\Delta}$.  Let $B$ be any set.  Then
+\begin{enumerate}
+\item $\displaystyle B \cup \left(\bigcap_{\alpha\in\Delta}A_{\alpha}\right)=\bigcap_{\alpha\in\Delta}(B\cup A_{\alpha})$,
+\item $\displaystyle B \cap \left(\bigcup_{\alpha\in\Delta}A_{\alpha}\right)=\bigcup_{\alpha\in\Delta}(B\cap A_{\alpha})$.
+\end{enumerate}
+(You only need to prove one of these; the other is similar.)
+\end{theorem}
+
+\begin{theorem}[Generalized DeMorgan's Law]
+Suppose we have a collection $\{A_{\alpha}\}_{\alpha\in\Delta}$.  Then
+\begin{enumerate}
+\item $\displaystyle \left(\bigcup_{\alpha\in\Delta} A_{\alpha}\right)^C=\bigcap_{\alpha\in\Delta}A_{\alpha}^{C}$,
+\item $\displaystyle \left(\bigcap_{\alpha\in\Delta} A_{\alpha}\right)^C=\bigcup_{\alpha\in\Delta}A_{\alpha}^{C}$.
+\end{enumerate}
+(You only need to prove one of these; the other is similar.)
+\end{theorem}
+
+\end{section}

IntroToProof.tex

@@ -126,6 +126,9 @@
 \include*{EvenMoreQuantification}
 \include*{IntroSetTheoryTopology}
 \include*{Sets}
+\include*{PowerSetsParadoxes}
+\include*{IndexingSets}
+\include*{Topology}
 %Appendices
 \include{ElementsOfStyle}
 \include{FancyMathematicalTerms}

PowerSetsParadoxes.tex

@@ -0,0 +1,121 @@
+\begin{section}{Power Sets and Paradoxes}
+
+We've already seen that using union, intersection, set difference, and complement we can create new sets (in the same universe) from existing sets.  In this section, we will describe another way to generate new sets; however, the new sets will not ``live" in the same universe this time.
+
+\begin{definition}
+If $S$ is a set, then the \textbf{power set} of $S$ is the set of subsets of $S$.  The power set of $S$ is denoted $\mathcal{P}(S)$.
+\end{definition}
+
+\begin{remark}
+It follows immediately from the definition that $A\subseteq S$ iff $A\in\mathcal{P}(S)$.\footnote{Recall that ``iff" is an abbreviation for `if and only if", which is a statement of the form $A\iff B$ for propositions $A$ and $B$.  Recall that this is short for both $A\implies B$ \emph{and} $B\implies A$.}  It is important to pay close attention to whether ``$\subseteq$" or ``$\in$" is the proper symbol to use.
+\end{remark}
+
+\begin{example}
+If $S=\{a,b\}$, then $\mathcal{P}=\{\emptyset, \{a\}, \{b\}, S\}$.
+\end{example}
+
+\begin{question}
+Implicit in the definition of power set is that $S$ is a subset of some fixed universe $U$.  What universe does it make sense for $\mathcal{P}(S)$ to live in?
+\end{question}
+
+\begin{exercise}
+For each of the following sets, find the power set.
+\begin{enumerate}
+\item $W=\{\circ, \triangle, \square\}$
+\item $O=\{a,\{a\}\}$
+\item $R=\emptyset$
+\item $D=\{\emptyset\}$
+\end{enumerate}
+\end{exercise}
+
+\begin{conjecture}
+How many subsets do you think that a set with $n$ elements has?  What if $n=0$?  You do not need to prove your conjecture at this time.  We will prove this later using mathematical induction.
+\end{conjecture}
+
+\begin{exercise}
+Do your best to describe $\mathcal{P}(\mathbb{N})$.  You cannot write down all of $\mathcal{P}(\mathbb{N})$.  Why not?
+\end{exercise}
+
+\begin{remark}
+It is important to realize that the concepts of \emph{element} and \emph{subset} need to be carefully delineated.  For example, consider the set $A=\{x,y\}$.  The object $x$ is an element of $A$, but the object $\{x\}$ is both a subset of $A$ and an element of $\mathcal{P}(A)$.  This can get confusing rather quickly.  Consider the set $O$ from the previous example.  The set $\{a\}$ happens to be an element of $O$, a subset of $O$, and an element of  $\mathcal{P}(O)$.
+\end{remark}
+
+\begin{theorem}
+Let $S$ and $T$ be sets.  Then $S\subseteq T$ iff $\mathcal{P}(S)\subseteq \mathcal{P}(T)$.\footnote{To prove this theorem, you have to write two distinct subproofs: $A\implies B$ and $B\implies A$.}
+\end{theorem}
+
+\begin{theorem}
+Let $S$ and $T$ be sets.  Then $\mathcal{P}(S)\cap\mathcal{P}(T)=\mathcal{P}(S\cap T)$.
+\end{theorem}
+
+\begin{theorem}
+Let $S$ and $T$ be sets.  Then $\mathcal{P}(S)\cup\mathcal{P}(T)\subseteq \mathcal{P}(S\cup T)$.
+\end{theorem}
+
+
+\begin{exercise}
+Let $S$ and $T$ be sets.
+\begin{enumerate}
+\item Provide a counterexample to show that it is not necessarily true that $\mathcal{P}(S)\cup\mathcal{P}(T)= \mathcal{P}(S\cup T)$.
+\item Is it ever true that $\mathcal{P}(S)\cup\mathcal{P}(T)= \mathcal{P}(S\cup T)$ or are $\mathcal{P}(S)\cup\mathcal{P}(T)$ and $\mathcal{P}(S\cup T)$ always different sets?
+\end{enumerate}
+\end{exercise}
+
+We now turn out attention to the issue of whether there is one mother of all universal sets.  Before reading any further, consider this for a moment.  That is, is there one largest set that all other sets are a subset of?  Or, in other words, is there a set of all sets?  To help wrap our heads around this issue, consider the following riddle, known as the \textbf{Barber of Seville Paradox}.
+
+\begin{quote}
+In Seville, there is a barber who shaves all those men, and only those men, who do not shave themselves.  Who shaves the barber?
+\end{quote}
+
+\begin{problem}\label{barber}
+Discuss the Barber of Seville Paradox.  Does the barber shave himself or not?
+\end{problem}
+
+Problem~\ref{barber} is an example of a \textbf{paradox}.  I haven't defined paradox.  What do you think it means?  Now, suppose that there is a set of all sets and call it $\mathcal{U}$.  Then we can write $\mathcal{U}=\{A:A\mbox{ is a set}\}$.
+
+\begin{problem}
+Given our definition of $\mathcal{U}$, explain why it is an element of itself.
+\end{problem}
+
+If we continue with this line of reasoning, it must be the case that some sets are elements of themselves and some are not.  Let $X$ be the set of all sets that are elements of themselves and let $Y$ be the set of all sets that are not elements of themselves.
+
+\begin{question}
+Does $Y$ belong to $X$ or $Y$?  Explain why this is a paradox.
+\end{question}
+
+The above paradox is one way of phrasing a paradox referred to as \textbf{Russell's paradox}.  Okay, how did we get into this mess in the first place?!  By assuming the existence of a set of all sets, we can produce all sorts of paradoxes.  The only way to avoid the paradoxes is to conclude that there is no set of all sets.  Here is some more evidence that we shouldn't assume the existence of a set of all sets.
+
+\begin{question}
+If $\mathcal{U}$ is the set of all sets, then what is the relationship between $\mathcal{U}$ and $\mathcal{P}(\mathcal{U})$?  What about $\mathcal{P}(\mathcal{P}(\mathcal{U})$?
+\end{question}
+
+The upshot is that the collection of all sets is \emph{not} a set!  Here are some additional paradoxes.
+
+\begin{problem}
+Pick any two of the paradoxes below and for each one explain why it is a paradox.
+\end{problem}
+
+%The following paradoxes are from Dave Richeson.
+\noindent \textbf{Librarian's Paradox.} A librarian is given the unenviable task of creating two new books for the library. Book A contains the names of all books in the library that reference themselves and Book B contains the names of all books in the library that do not reference themselves. But the librarian just created two new books for the library, so their titles must be in either Book A or Book B. Clearly Book A can be listed in Book B, but where should the librarian list Book B?\\
+
+\noindent \textbf{Liar's Paradox.} Consider the statement: this sentence is false. Is it true or false?\\
+
+\noindent \textbf{Berry Paradox.} Consider the claim: every natural number can be unambiguously described in fourteen words or less. It seems clear that this statement is false, but if that is so, then there is some smallest natural number which cannot be unambiguously described in fourteen words or less. Let's call it $n$. But now $n$ is ``the smallest natural number that cannot be unambiguously described in fourteen words or less.'' This is a complete and unambiguous description of $n$ in fourteen words, contradicting the fact that $n$ was supposed not to have such a description. Therefore, all natural numbers can be unambiguously described in fourteen words or less!\\
+
+\noindent \textbf{The Naming Numbers Paradox.} Consider the claim: every natural number can be unambiguously described using no more than 50 characters (where a character is a--z, 0--9, and a ``space''). For example, we can describe 9 as ``9'' or ``nine'' or ``the square of the second prime number.'' There are only 37 characters, so we can describe at most $37^{50}$ numbers, which is very large, but not infinite. So the statement is false. However, here is a ``proof'' that it is true. Let $S$ be the set of natural numbers that can be unambiguously described using no more than 50 characters. For the sake of contradiction, suppose it is not all of $\mathbb{N}$. Then there is a smallest number $t\in\mathbb{N}-S$. We can describe $t$ as: the smallest natural number not in $S$.  Thus $t$ can be described using no more than 50 characters. So $t\in S$, a contradiction.\\
+
+\noindent \textbf{Euathlus and Protagoras.} Euathlus wanted to become a lawyer but could not pay Protagoras. Protagoras agreed to teach him under the condition that if Euathlus won his first case, he would pay Protagoras, otherwise not. Euathlus finished his course of study and did nothing. Protagoras sued for his fee. He argued:\\
+
+\noindent If Euathlus loses this case, then he must pay (by the judgment of the court).\\
+If Euathlus wins this case, then he must pay (by the terms of the contract).\\
+He must either win or lose this case.\\
+Therefore Euathlus must pay me.\\
+
+\noindent But Euathlus had learned well the art of rhetoric. He responded:\\
+
+\noindent If I win this case, I do not have to pay (by the judgment of the court).\\
+If I lose this case, I do not have to pay (by the contract).\\
+I must either win or lose the case.\\
+Therefore, I do not have to pay Protagoras.
+
+\end{section}