Homework 2

Solution.

Solution 1 (using open sets)

This is the solution suggested by the hint. Note that $V_n = K_1 \cap \p{X \setminus K_n}$. Since $K_n$ is compact, it's automatically closed, so $X \setminus K_n$ is open. Thus, $V_n$ is relatively open in $K_1$. Following the hint, we assume that $\bigcap_{n=1}^\infty K_n = \emptyset$, which implies

In other words, $\set{V_n}_{n \in \N}$ is an open cover of $K_1$, so because $K_1$ is compact, there exist $n_1 < n_2 < \cdots n_k$ such that $\set{V_{n_1}, \ldots, V_{n_k}}$ is a finite subcover of $K_1$. Since $K_n \supseteq K_{n+1}$ for all $n$, we have $V_n \subseteq V_{n+1}$ for all $n$. Thus,

But this implies that $K_{n_k} = \emptyset$, which is impossible.

Solution 2 (using sequences)

For each $n \in \N$, we let $x_n \in K_n$ and consider the sequence $\p{x_n}_n \subseteq K_1$. Since $K_1$ is compact, there exists a subsequence $\p{x_{n_k}}_k$ which converges to some $x_0 \in K_1$. Since $\p{n_k}_k$ is a subsequence, we have $n_k \geq k$ for each $k$. Thus, given $\ell \in \N$, because $K_n \supseteq K_{n+1}$ for all $n$, we get

In other words, $\p{x_{n_k}}_{k=\ell}^\infty \subseteq K_\ell$ and $x_{n_k} \to x_0$, so because compact sets are closed, we get $x_0 \in K_\ell$. Since $\ell$ was arbitrary, this shows

i.e., the intersection is non-empty.

Solution.

1. "$\implies$"

   We have already proven that compact sets are closed (this is Corollary 1.5.6).

   $\impliedby$

   Assume $Z$ is closed and let $\p{z_n}_n \subseteq Z$ be a sequence. Then $\p{z_n}_n \subseteq Y$, so because $Y$ is compact, there exists a subsequence $\p{z_{n_k}}_k$ which converges to some $y_0 \in Y$. But $\p{z_{n_k}}_k \subseteq Z$ and $z_{n_k} \to y_0$, so because $Z$ is closed, we have $y_0 \in Z$. Thus, $Z$ is compact.

2. Let $\set{U_\alpha}_{\alpha \in I}$ be an open cover of $Y_1 \cup \cdots \cup Y_n$. Notice that for any $1 \leq i \leq n$,

   $$Y_i \subseteq \bigcup_{j=1}^n Y_j \subseteq \bigcup_{\alpha \in I} U_\alpha.$$

   In other words, $\set{U_\alpha}_{\alpha \in I}$ is also an open cover of each $Y_i$, which is compact, so there exists a finite subset $F_i \subseteq I$ such that $\set{U_\alpha}_{\alpha \in F_i}$ is an open cover of $Y_i$. Then

   $$\bigcup_{i=1}^n Y_i \subseteq \bigcup_{i=1}^n \bigcup_{\alpha \in F_i} U_\alpha,$$

   i.e., $\set{U_\alpha \mid \alpha \in F_1 \cup \cdots \cup F_n}$ is an open cover of the union, and it is finite since each $F_i$ is a finite set.

3. Let $E \subseteq X$ be a finite subset and $\set{U_\alpha}_{\alpha \in I}$ be an open cover of $E$. If $E = \emptyset$, then $E \subseteq U_\alpha$ for any $\alpha \in I$, so $E$ is compact.

   Next, if $E = \set{x}$ is a singleton set, then because $\set{U_\alpha}_{\alpha \in I}$ covers $E$, there exists $\alpha_0 \in I$ such that $x \in U_{\alpha_0}$. Then $E \subseteq U_{\alpha_0}$, so $E$ is compact.

   Finally, if $E = \set{x_1, \ldots, x_n}$, then

   $$E = \bigcup_{i=1}^n \set{x_i}.$$

   Applying (ii) with $Y_i = \set{x_i}$, which we just proved is compact, shows that $E$ is also compact.

Solution.

Let $\varepsilon > 0$. Since $g$ is uniformly continuous, there exists $\eta > 0$ such that

Similarly, because $f$ is uniformly continuous, there exists $\delta > 0$ such that

Putting it all together, if $d_X\p{x, x'} < \delta$, then by ($\dagger$), $d_Y\p{f\p{x}, f\p{x'}} < \eta$, so applying ($*$) with $\p{y, y'} = \p{f\p{x}, f\p{x'}}$, we get $d_Z\p{g\p{f\p{x}}, g\p{f\p{x'}}} < \varepsilon$.