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\title{18.100C Writing Assignment 4}
\author{Tony Kim}
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In this assignment we are to discuss either compactness or connectedness. I will give an alternative proof that compact sets are closed (Rudin 2.34); and solve Problem 6 in Rudin Ch. 4.

\section{Compact subsets of metric spaces are closed.}

Suppose not. Let $K$ be a compact subset of a metric space $X$. We assume that $K$ is not closed.

Since $K$ is not closed, there exists a limit point $x$ that lies in $K^c$. We will construct an open cover of K by the following. For each integer $n = 1, 2, 3,...$, we define:

\begin{equation}
	V_n = \left\{ p \in K \ | \ p \notin N_{1/n}(x) \right\}
\end{equation}

where $N_{1/n}(x)$ is as usual the neighborhood of radius $1/n$ about $x$.

Clearly $\bigcup_{n}V_n$ is an open cover of $K$. However it has no finite subcover. Since $V_{n+1} \supset V_n$, a finite union $V_{n_1} \cup V_{n_2} \cup ... \cup V_{n_k} = V_M$, where $M = max(n_1, n_2,..., n_k)$. But since $x$ is a limit point of $K$, the set $N_{1/M}(x) \cap K$ is nonempty; and hence the finite union does not cover $K$. Contradiction.

\section{Rudin Ch 4, Problem 6.}

\subsection{If $f$ is defined on $E$, the \emph{graph} $G$ of $f$ is the set of points $(x, f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, $G$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.}


We will first show that the continuity of $f$ and the compactness of $E$ imply that $G$ is a bounded and closed subset of $\Re^2$; and hence that it is compact.

Since $f$ is continuous and $E$ is compact, the image $f(E)$ is also compact (Rudin 4.14). Furthermore, both $E$ and $f(E)$ are subsets of $\Re$; hence by Heine-Borel they are closed and bounded. Let $x_1, x_2, y_1, y_2$ be such that $x_1 < x < x_2$ for every $x \in E$ and $y_1 < y < y_2$ for every $y \in f(E)$.

The graph $G$ consists of points $(x, f(x))$ for $x \in E$. Hence, by the above inequalities, $G$ is contained in the rectangular subset of the plane $\left\{(x,y) \in \Re^2 \ | \ x_1 < x < x_2; \ y_1 < y < y_2 \right\}$. Clearly, $G$ is bounded.

Now we show that $G$ is closed. Suppose $p = (a,b)$ is a limit point of $G$. So for any neighborhood about $p$, we can find a point $(x,f(x))$ in $G$. In particular, this implies that $a$ is a limit point of $E$. Since $E$ is closed, $a$ lies in $E$. We now show that it must be the case that $b = f(a)$.

Assume not. Suppose $D = d((a,b), (a,f(a))) > 0$. Since $f$ is continuous, for $\epsilon = \frac{D}{2}$ there exists some corresponding $\delta$ such that whenever $d(x,a)<\delta$ for any $x \in E$, we have $d(f(x),f(a))<\frac{D}{2}$. Choose $r = min(\frac{D}{2}, \delta)$; the neighbordhood $N_r((a,b))$ then has no points of $G$. To show this, we consider the interval $(a-r, a+r)$ only, since any point that lies in $N$ has an x-coordinate within this segment. But since $r \leq \delta$, for every $x$ in the interval we have $f(x) < f(a) + \frac{D}{2}$. On the other hand, any point that lies in $N$ has a y-coordinate $y$ that satisfies $y > f(a)-r \geq f(a)-\frac{D}{2}$. No point can satisfy both inequalities and hence $N \cap G$ is empty. Contradiction.

Thus $G$ is a closed and bounded subset of $\Re^2$. Heine-Borel applies and $G$ is compact.

To prove the converse, let $G$ be compact but suppose that $f$ is not continuous. Then there is some point $p \in E$ such that $\lim_{x \to p} f(x) \neq f(p)$. 

Recall that for $\lim_{x \to p} f(x) \neq f(p)$, $p$ must be a limit point of $E$. If $p$ were instead an isolated point, we can find a $\delta$ such that $N_\delta(p) \cap E = \left\{p\right\}$. Then, for any $\epsilon > 0$, we have for every $x \in E$ for which $d(x,p)<\delta$ that $d(f(x),f(p)) < \epsilon$, simply because the only point in $E$ that satisfies the criterion is $x = p$. So if $p$ is not a limit point, then we trivially have $\lim_{x \to p} f(x) = p$, contrary to our condition.

For $\lim_{x \to p} f(x) \neq p$, there are two possibilities:

Suppose $q = \lim_{x \to p} f(x)$ exists but $q \neq f(p)$. Here, we will argue that $(p,q)$ is a limit point of $G$ but does not lie in it. So $G$ is not closed and therefore cannot be compact.

Fix $\epsilon > 0$. Since $\lim_{x \to p} f(x) = q$, there exists a $\delta > 0$ such that for every $x \in E$ that satisfies $d(x,p)<\delta$, we have $d(f(x),f(p)) < \frac{\epsilon}{2}$. Put $r = min(\delta, \frac{\epsilon}{2})$. Since $p$ is a limit point of $E$, we are able to select a point $x$ from $N_r(p) \cap E$. Then, by the triangle inequality:
\begin{eqnarray*}
	d((x,f(x)),(p,q)) &\leq& d((x,f(x)),(p,f(x))) + d((p,f(x)),(p,q))\\
										&<&    \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{eqnarray*}

Since $(x,f(x)) \in G$, $(p,q)$ is a limit point of $G$. However $(p,q)$ cannot lie in $G$ since the only element in $G$ with $p$ as its x-coordinate is $(p,f(p)) \neq (p,q)$. So $G$ is not closed, and hence it cannot be compact. Contradiction.

Now we consider the case when $\lim_{x \to p} f(x)$ does not exist. Again, we take advantage of the fact that $p$ is a limit point in $E$.

First, we define the ``lower'' and ``upper'' sequences $x^{-}_k$ and $x^{+}_k$ in $E$ by the following.

For $n = 1,2,3,...$, define $x^{-}_n$ by choosing an element from the set $(p-\frac{1}{n}, p) \cap E$. Likewise, we define $x^{+}_n$ from $(p, p+\frac{1}{n}) \cap E$. Since the sequences are in $E$, $f(x^{-/+}_k)$ is defined and we can consider the sequences in $\Re^2$:
\begin{eqnarray*}
	G^{-}_k &=& (x^{-}_k,f(x^{-}_k))\\
	G^{+}_k &=& (x^{+}_k,f(x^{+}_k))
\end{eqnarray*}

By construction $\left\{G^{-}_k\right\}$ and $\left\{G^{+}_k\right\}$ are infinite subsets of $G$; and, by sequential compactness, must have limits in $G$. Furthermore, it is clear that the x-coordinate of both limits must be $p$. Let $G^{-}_k \rightarrow (p,y^{-})$ and $G^{+}_k \rightarrow (p,y^{+})$. Since $G$ is compact, it is closed and both points must lie in $G$.

However, $(p,y^{-})$ must be distinct from $(p,y^{-})$. If instead $q = y^{-} = y^{+}$, then for any $\epsilon > 0$ we can find $\delta^{-}$ such that $d(x,p)<\delta^{-}$ with $x \in (p-1, p) \cap E$ implies that $d(f(x),f(p))< \epsilon$. Similarly, we can obtain a $\delta^{+}$ for the upper seqeunce. If we let $\delta = min(\delta^{-}, \delta{+})$, we see that $\lim_{x \to p} f(x) = (p,q)$, contrary to our premise that the limit does not exist.

So, $(p,y^{-})$ and $(p,y^{+})$ are in $G$ and they are distinct. This contradicts the notion that any function assigns only one value to an element in its domain. 

Therefore, the compactness of $G$ implies the continuity of $f$.

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