In Stein's real analysis Page $ 188 $ , the author states that :
Definition $1$ : $H$ is a Hilbert space . A set $X \subset H$ is compact is for any sequence $\{f_n \}$ in $X$ , there exists a subsequence $\{ f_{n_k}\}$ that converges in the norm to an element in $X$ .
This definition of a compact set is quite different from what I have learned :
Definition $2$: A set $X \subset H$ is compact if every open covering of $X$ has a finite subcovering.
I want to show that the two definitions of the compact set above are the same.
My attempt :
First I want to use $2$ to prove $1$ . Since $H$ is a Hilbert space , then it must be a $T_1$ space , so $X$ is closed . Let $X'$ denote the closure of $\{f_n \}$ , so $X' \subset X$ is also a compact set . And the collection of ball $B_{2^{-n}}(f_n)$ covering $X'$ , so that by $2$ , we have a finite subcovering . Notice that there are infinite many elements of $\{f_n\}$ in some ball $B_{2^{-N}}(f_N)$ . Let $f_{n_1} = f_N$ and we say that $f_{k_2},…,f_{k_n},..$ are the elements in $B_{2^{-N}}(f_{n_1})$ . We can proceed by the same way to find a sequence $f_{n_1} ,…, f_{n_k},…$ and they form a Cauchy sequence . By the completeness of Hilbert space , we then conclude the proof .
The question is how to use $1$ to prove $2$ . I have seen a hint here . Suppose there is family of open sets $O_1,\ldots,O_n,\ldots$ such that $X \subset \cup O_n$ but there is no finite subcovering. Construct the sequence $(x_n)$ such that $x_n \in X \setminus \bigcup_{i=1}^n O_i $ , then we can find a subsequence $\{x_{n_k} \}$ and for some $N$ , each $x_{n_k} \notin O_N$ while $\{x_{n_k} \}$ converges to an element $x \in O_N$ . But how to deal with this next ?
For the proof above, we see that we only assume $H$ is a complete metric space. Does the conclusion valid when $H$ is a metric space (might not be complete).
Or more generally , does the conclusion still valid when $H$ is just a topology space . Here definition $1$ is :
for any sequence $\{ f_n\}$ in $X$ , there exist $f \in X$ such that for every open set containing $f$ , it also contains at least one element $f_k$
Best Answer
To finish your argument for (1) implies (2), we pick $x_n \in X\setminus \cup_{i=1}^n O_i$, then let $(y_n)$ be a subsequence of $(x_n)$ which converges to $y\in X$. Since the $O_i$ are an open cover of $X$, there is some $N$ so that $y\in O_N$. But there are infinitely many $y_n \notin O_N$, which is impossible: if $(y_n)\to y$, then for any open neighborhood $U$ of $y$, there must be some $m$ so that for $n\ge m$, we have $y_n\in U$.
To prove that (2) implies (1) without completeness, let me define a limit point of $Y=\{f_n\}$: a point $z\in H$ is a limit point of $Y$ if for every open neighborhood $U$ of $z$, we have that $(U\setminus \{z\}) \cap Y \ne \emptyset$. It's easy to show that in any topological space, the closure of a set $Y$ is equal to $Y$ along with the limit points of $Y$. Moreover, if $z$ is a limit point of $Y = \{f_n\}$, then it's easy to show that there is a subsequence of $(f_n)$ which converges to $z$: pick $n_1\in\mathbb{N}$ so that $f_{n_1} \in (B_{1}(z)\setminus \{z\}) \cap Y$, then let $\delta_1 = \mbox{dist}(z,f_{n_1})$ and pick $n_2\in \mathbb{N}$ so that $f_{n_2} \in (B_{\delta_1/2}(z)\setminus \{z\}) \cap Y$, and so on. This subsequence $(f_{n_k})$ converges to $z$. So now, as you say, this closure, which you called $X'$, is compact. If $X'$ contains any limit points of $Y$, then we're done. Otherwise, $X' = Y$ and there are no limit points of $Y$. Thus, for each $n\in\mathbb{N}$, there is an open neighborhood $U_n$ of $f_n$ with $Y\cap U_n = \{f_n\}$. Now the $U_n$ form an open cover of $Y$ with no finite subcover.
You're right! We didn't use much about the Hilbert space $H$. It's true for all metric spaces. Though you should change your second definition of compactness to
It's not true for all topological spaces. Not even for Hausdorff topological spaces.