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\homework{11: Introduction to Analysis}{Due Friday, Week 6}
This problem set relies on several concepts we talked about in class! We list them here:
\begin{itemize}
\item \textbf{Limits}: We say that a sequence $\{a_n\}_{n=0}^\infty$ has the limit $L$ as $n$ approaches infinity if ``the terms of $\{a_n\}_{n=0}^\infty$ go to $L$ as $n$ gets large.'' Formally, we say 
\begin{align*}
\lim_{n \to \infty} a_n = L \Leftrightarrow \forall \epsilon >0, \exists N \forall n > N, |a_n - L| < \epsilon.
\end{align*}
Essentially, for any amount of ``close to $L$'' we want, there is some $N$ past which if we pick $a_n$'s with $n > N$, we have that $a_n$ is as close to $L$ as we wanted.
\item \textbf{Cauchy}: We say that a sequence $\{a_n\}_{n=0}^\infty$ is Cauchy if ``the terms of $\{a_n\}_{n=0}^\infty$ get close to each other as $n$ gets large.'' Formally, we say 
\begin{align*}
\{a_n\}_{n=0}^\infty \textrm{ is Cauchy} \Leftrightarrow \forall \epsilon >0, \exists N \forall m,n > N, |a_n - a_m| < \epsilon.
\end{align*}
Essentially, for any amount of ``close to $L$'' we want, there is some $N$ past which if we pick $a_m, a_n$'s with $m,n > N$, we have that $a_n$ is as close to $a_m$ as we wanted.  It's \textit{like} having a limit, except that you don't actually know where the sequence is going; just that the terms are getting close to each other!
\item \textbf{Measure}: We say that any function $\mu$ that takes subsets of $\mathbb{R}$ to $\mathbb{R}^{\geq 0}$ is a \textbf{measure}\footnote{Strictly speaking, a Lebesgue measure.} if it satisfies the following properties:
\begin{enumerate}
\item \textbf{Base cases}: $\mu(\emptyset) = 0$, $\mu([0,1]) = 1$.
\item \textbf{Translation/Scaling}:  For any set $A$ that $\mu$ is defined on and any $x \in \mathbb{R}, \lambda \in \mathbb{R}^+$, we have $\mu(A) = \mu(A+x)$ and $\mu(\lambda \cdot A) = \lambda \cdot \mu(A)$.
\item \textbf{Subadditive}: If $A \subseteq B$ are sets that $\mu$ is defined on, then $\mu(A) \leq \mu(B)$.
\item \textbf{Additive}: If $\{A_i\}_{i=1}^\infty$ is a collection of disjoint sets that $\mu$ is defined on, then $\sum_{i=1}^\infty \mu(A_i) = \mu \left( \bigcup_{i=1}^n A_i \right)$.
\end{enumerate}
\end{itemize}
Here are the problems!  Do \textbf{three} of the following \textbf{six}.
\begin{enumerate}
\item Consider the following sequence:
\begin{align*}
a_1 = 1, a_n = a_{n-1} + \frac{1}{n^2}.
\end{align*}
Show that $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence.


\item When we defined the real numbers using Cauchy sequences in class, many people asked why we couldn't just use the definition of ``sequences with limits'' in place of the stranger ``Cauchy'' definition.  This problem explains why we used Cauchy!  Specifically, consider the following ``alternate'' construction for the real numbers:
\begin{itemize}
\item Let $S$ be the set of all sequences $\{a_n\}_{n=1}^\infty$ of rational numbers with the following property: there is some $L \in \mathbb{Q}$ such that $\lim_{n \to\infty} a_n = L$.
\item Call two sequences with limits \textbf{equivalent} if they have the same limit: formally, say that $\{x_n\}_{n=1}^\infty \sim \{y_n\}_{n=1}^\infty$ if and only if there is some $L \in \mathbb{Q}$ such that $L = \lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$.
\item Define the real numbers as $S/ \sim$, i.e.\ the collection of all equivalence classes of sequences of rational numbers under $\sim$.
\end{itemize}
Explain why this construction does not give you $\mathbb{R}$. 


\item Take our rules for being a measure, and remove the ``translation/scaling'' property: call such an object a ``meamaybe.''  Find a meamaybe that assigns 0 or 1 to every subset of $\mathbb{R}$, and explain why it satisfies the three properties of being a meamaybe.


\item A set $S \subseteq \mathbb{R}$ has an \textbf{upper bound} if there is some element $U \in \mathbb{R}$ such that 
\begin{align*}
\forall s \in S, ~U \geq s.
\end{align*}
Similarly, a set $S \subseteq \mathbb{R}$ has a \textbf{lower bound} if there is some element $F \in \mathbb{R}$ such that 
\begin{align*}
\forall s \in S,~ F \leq s.
\end{align*}
Show that if a sequence $\{ a_n\}_{n = 1}^\infty$ of real numbers has a limit $L \in \mathbb{R}$, then it has an upper and lower bound.


\item Show that if a sequence $\{a_n\}_{n=1}^\infty$ has a limit, then it is Cauchy.


\item  For any interval $[a,b]$, we define the process of removing the ``middle-$n$-th'' as simply removing the open interval 
\begin{align*}
\left( \frac{b+a}{2} - \frac{b-a}{2n}, \frac{b+a}{2} + \frac{b-a}{2n}\right).
\end{align*}
from this set.  For example, removing the middle-$3$-rd from $[0,1]$ is just removing
\begin{align*}
\left( \frac{1}{2} - \frac{1}{6}, \frac{1}{2} + \frac{1}{6} \right) = \left( \frac13, \frac23\right)
\end{align*}
from $[0,1]$, which leaves us with the set $[0,1/3] \cup [2/3, 1]$.  Note that if we started with a closed interval, this always leaves us with two closed intervals.
The Cantor set $C_\infty$ was created iteratively, by setting $C_0  =[0,1]$ and saying that if $C_n$ was defined, we can make $C_{n+1}$ by removing the middle-third from each interval in $C_{n}$.  From there, we defined $C_\infty$ as the intersection of all of these sets.

Consider the following ``Cantor-like'' set $S_n$:
\begin{itemize}
\item $S_0 = [0,1]$.
\item If we have constructed $S_{n-1}$, create $S_n$ out of it by removing the middle $\frac{1}{4^n}$-th from each interval in $S_{n-1}$.
\end{itemize}
Set $S_\infty$ as the intersection of all of these sets.  What is the measure of $S_n$?
\end{enumerate}


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