AS ANA2 2

Problem 1

Suppose for every $x\in S$, there exists a neighborhood $U_s$ such that $U_s\cap S$ is countable. $\{U_s{\}}_{s\in S}$ is an open cover of $S$. Since every subset of a separable metric space is separable, $S$ is separable. By Lindelöf’s covering theorem, there exists a countable subcover $\{U_s{\}}_{s\in \alpha \subset S}$. This implies

$$|S|=\left|\bigcup _{s\in \alpha }{U}_s\cap S\right|={\aleph }_0,$$

a contradiction.

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Problem 2

Denote the graph of $f$ by $G$. Suppose $f$ is continuous. Define $g:X\to X\times Y$ by $x\mapsto (x,f(x))$. Since its components are continuous, $g$ is continuous. The image of a compact set under a continuous map is compact, so $G$ is compact. In particular, $G$ is closed.

Conversely, suppose $G$ is closed. Let $p$ be a limit point of $X$. Let $\{p_n\}\subseteq X$ be a sequence converging to $p$. Consider the sequence $\{g(p_n)\}\subseteq X\times Y$. Since $X\times Y$ is compact, it is limit point compact, so $\{g(p_n)\}$ has a limit point $(a,b)$, which must then be in $G$ because $G$ is closed. Let $\{q_n\}\subseteq \{g(p_n)\}$ be a sequence converging to $(a,b)$. The component sequences of $\{q_n\}$ in $X$ and $Y$ must converge to $a$ and $b$. Since $p$ is the only limit point of $\{p_n\}$, it follows that $a=p$. Since $f$ is a function, this forces $b=f(p)$. Thus, $\{g(p_n)\}$ has exactly one limit point, and it is $(p,f(p))$. It follows that $\{g(p_n)\}\to (p,f(p))$. Therefore, $\{f(p_n)\}\to f(p)$ and $f$ is continuous at $p$.

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Problem 3

If $X$ is compact and $f:X\to \mathbb{R}$ is any continuous function, $f(X)$ is compact, and in particular bounded.

Assume every real valued function on $X$ is bounded. Suppose $X$ is not compact. Then $X$ is not sequentially compact. Let $S=\{x_n{\}}_{n=1}^{\infty }\subseteq X$ not have a limit point. For each $x_n$, there exists $r_n$ such that $B_{r_n}(x_n)\cap S$ is the singleton $\{x_n\}$ and all of $\{B_{r_n}(x_n)\}$ are disjoint from each other.

For each natural number $n$, define $f_n$ by

$$f_n=\{\begin{array}{ll}1-2d(x,x_n)/r_n& d(x,x_n)\le r_n/2\\ 0& \text{otherwise}.\end{array}$$

Define the function $f:X\to \mathbb{R}$ by

$$f(x)=\sum _{n=1}^{\infty }n{f}_n(x)\text{ for all }x\in X.$$

Since each $f_n$ is continuous and vanishes outside $B_{r_n/2}(x_n)$ and the collection $\{B_{r_n}(x_n)\}$ is disjoint, $f$ is well-defined and continuous. But for each natural number $n$, $f(x_n)=n$, and hence $f$ is unbounded above. This is a contradiction.

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Problem 4

Let $S\subseteq Z$ be closed. It suffices to prove that $g^{-1}(S)$ is closed in $Y$. Since $g\circ f$ is continuous, we know that $f^{-1}(g^{-1}(S))$ is closed in $X$. Since $X$ is compact, $f^{-1}(g^{-1}(S))$ is compact. Since $f$ is onto, $f(f^{-1}(g^{-1}(S)))=g^{-1}(S)$, which must be compact since $f$ is continuous. Since $Y$ is a metric space, this implies $g^{-1}(S)$ is closed.

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Problem 5

It is clear that

$$f\left(\bigcap _{n=1}^{\infty }{K}_n\right)\subseteq \bigcap _{n=1}^{\infty }f(K_n).$$

To prove the reverse inclusion, consider $l\in {\bigcap }_{n=1}^{\infty }f(K_n)$. For every $n$, there exists $x_n\in K_n$ such that $f(x_n)=l$. Consider set $\{x_n\}$ in $K_1$. Since $K_1$ is compact, it must have a convergent subsequence $\{y_n{\}}_{n=1}^{\infty }\to p\in K_1$. For every $k$, $\{y_n{\}}_{n=k}^{\infty }\subseteq K_k$, so $p$ is a limit point of and hence must be contained in $K_k$. Thus, $p\in {\bigcap }_{n=1}^{\infty }{K}_n$. Since $f$ is continuous, $f(p)={\lim }_{n\to \infty }f(x_n)=l$. Thus, $l\in f\left({\bigcap }_{n=1}^{\infty }{K}_n\right)$.