# $\mathcal{H}$ is relatively compact iff every sequence in $\mathcal{H}$ has a convergent subsequence?

I'm trying to prove that

Let $(Y,\rho_{Y}),(K,\rho_{K})$ a complete metric space and a compact metric space, respectively. Let, as well, $Z=\mathcal{C}^{0}(K,Y)$ the metric space of continuous fuctions, such that $K\longrightarrow Y$, with the uniform metric and $\mathcal{H}\subset Z$. $\mathcal{H}$ is relatively compact in $Z$ iff for every $\{f_{k}\}$, such that $f_{k}\in\mathcal{H}$, there is a subsequence $\{f_{k_{j}}\}$ that converges in $Z$, that is $f_{k_{j}}\rightarrow\varphi\in Z$.

Suppose $\mathcal{H}$ is relatively compact in $Z$. Then $\overline{\mathcal{H}}$ is compact. Therefore, every sequence in $\overline{\mathcal{H}}$ has a subsequence that converges in $\overline{\mathcal{H}}$. That, particularly, means that every sequence in $\mathcal{H}$, that is a sequence in $\overline{\mathcal{H}}$, has a subsequence that converges in $\overline{\mathcal{H}}$. Then, for every $\{f_{k}\}$, such that $f_{k}\in\mathcal{H}$, there is a subsequence $\{f_{k_{j}}\}$ that converges in $Z$, that is $f_{k_{j}}\rightarrow\varphi\in\overline{\mathcal{H}} \subseteq Z$.

Conversely, suppose that for every $\{f_{k}\}$, such that $f_{k}\in\mathcal{H}$, there is a subsequence $\{f_{k_{j}}\}$ that converges in $Z$, that is $f_{k_{j}}\rightarrow\varphi\in Z$. We want to show that $\mathcal{H}$ is relatively compact, or what is the same, show that $\overline{\mathcal{H}}$ is compact.

Now $\overline{\mathcal{H}}\subset Z$, but $Z$ is complete and $\overline{\mathcal{H}}$ is closed (since closure is closed), therefore $\overline{\mathcal{H}}$ is complete. Then it only remains to show that $\overline{\mathcal{H}}$ is totally bounded, for if it is then $\overline{\mathcal{H}}$ is compact.

My question is how to show this using the hypothesis that every sequence has a subsequence that converges in $Z$?

Thanks to the answers of @BrianMScott and @BenjaminLim. In the next lines I complete the proof I was doing with their useful hints, comments are welcome

Proof: Suppose $\mathcal{H}$ is relatively compact in $Z$. Then $\overline{\mathcal{H}}$ is compact. Therefore, every sequence in $\overline{\mathcal{H}}$ has a subsequence that converges in $\overline{\mathcal{H}}$. That, particularly, means that every sequence in $\mathcal{H}$, that is a sequence in $\overline{\mathcal{H}}$, has a subsequence that converges in $\overline{\mathcal{H}}$. Then, for every $\{f_{k}\}$, such that $f_{k}\in\mathcal{H}$, there is a subsequence $\{f_{k_{j}}\}$ that converges in $Z$, that is $f_{k_{j}}\rightarrow\varphi\in\overline{\mathcal{H}} \subseteq Z$.

Conversely, suppose that for every $\{f_{k}\}$, such that $f_{k}\in\mathcal{H}$, there is a subsequence $\{f_{k_{j}}\}$ that converges in $Z$, that is $f_{k_{j}}\rightarrow\varphi\in Z$. We want to show that $\mathcal{H}$ is relatively compact, or what is the same, show that $\overline{\mathcal{H}}$ is compact.

Now $\overline{\mathcal{H}}\subset Z$, but $Z$ is complete and $\overline{\mathcal{H}}$ is closed (since closure is closed), therefore $\overline{\mathcal{H}}$ is complete. Then it only remains to show that $\overline{\mathcal{H}}$ is totally bounded, for if it is then $\overline{\mathcal{H}}$ is compact. But, for that, we need only to prove that $\mathcal{H}$ is totally bounded, since closure of a totally bounded set is totally bounded.

Suppose that $\mathcal{H}$ is not totally bounded, then exists $\epsilon>0$ such that there isn't a finite covering of balls, for $\mathcal{H}$, of the form $\{B_{g_{i},\epsilon}^{\rho_{\infty}}\}$ with $g_{i}\in\mathcal{H}$. Therefore, we can choose a $f_{1}\in\mathcal{H}$ and there will be a $f_{2}\in\mathcal{H}$ such that $f_{2}\notin B_{f_{1},\epsilon}^{\rho_{\infty}}$. In the same way there will be a $f_{3}\notin B_{f_{1},\epsilon}^{\rho_{\infty}}\cup B_{f_{2},\epsilon}^{\rho_{\infty}}$ and in general there will be $f_{k}\notin\bigcup_{i=1}^{k-1}B_{f_{i},\epsilon}^{\rho}$. Hence, we have defined, inductively, a sequence that there is $\epsilon>0$ such that $\rho(f_{k},f_{\ell})\geq\epsilon$ for every $k\neq\ell$, since if $\rho(f_{k},f_{\ell})<\epsilon$ for sufficiently large $k,\ell$ therefore $f(k)\in B_{f_{\ell},\epsilon}^{\rho_{\infty}}$ which contradicts the former construction. Therefore, every subsequence is not Cauchy and therefore none of them has the chance to converge. But by hypothesis, every sequence in $\mathcal{H}$ has a convergent subsequence, and supposing that $\mathcal{H}$ was not totally bounded has led us to contradiction. Therefore $\mathcal{H}$ is totally bounded in $Z$ and, hence, $\overline{\mathcal{H}}$. It proves that, since $\overline{\mathcal{H}}$ is complete and totally bounded, $\overline{\mathcal{H}}$ is compact. Finally, $\mathcal{H}$ is relatively compact.

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If $\mathcal{H}$ is totally bounded, then so is $\overline{\mathcal H}$, so it suffices to show that $\mathcal{H}$ is totally bounded. If not, there is some $\epsilon>0$ such that $\mathcal H$ has no finite $\epsilon$-net. That is, there is no finite subset $\mathcal F\subseteq\mathcal H$ such that for each $h\in\mathcal H$ there is some $f\in\mathcal F$ such that $\|h-f\|<\epsilon$. Use this to construct an infinite sequence $\langle h_n:n\in\Bbb N\rangle$ in $\mathcal H$ such that $\|h_m-h_n\|\ge\epsilon$ whenever $m\ne n$. Can this sequence have a subsequence that converges in $Z$?

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Oh! Great. I have not used definition of totally bounded, and that construction is very straightforward since I used something like that before. That's clear, I'll never forget that. Thanks. Answering your question: No, because every subsequence will not be Cauchy and therefore they will not converge. That is a contradiction with the hypothesis, therefore $\mathcal{H}$ is totally bounded. Thanks @BrianMScott. –  elessartelkontar May 21 '12 at 3:07

Let me try to expand a little on Brian's hint above. If $\mathcal{H}$ is not totally bounded, then there is $\epsilon > 0$ such that for no finite subcollection $f_1,\ldots, f_n$ of functions in $\mathcal{H}$ does the union

$$B_{\epsilon}(f_1) \cup \ldots \cup B_{\epsilon}(f_n)$$

cover $\mathcal{H}$. Now in particular this means the following: Choose some $f_1 \in \mathcal{H_1}$ and now choose $f_2 \in \mathcal{H}$ such that $||f_1 - f_2||\leq \epsilon$. Having chosen $f_1,\ldots,f_n \in \mathcal{H}$ it is always possible to choose $f_{n+1} \in \mathcal{H}$ such that

$$||f_{n+1} - f_i || \geq \epsilon$$

for $1 \leq i \leq n$. Now we consider the sequence $\{f_n\}$ that lives in $\mathcal{H}$. Suppose it has a convergent subsequence $\{f_{n_j}\}$ that converges to some $\varphi \in Z$. Now this also means that for our $\epsilon$ that we chose above, the ball of radius $\epsilon/2$ about $\varphi$ must contain all but finitely many terms of your subsequence. However you have a problem because such a ball can only contain at most $1$ term of your subsequence.

To see why, say you have two terms of the subsequence $f_{n_1}$ and $f_{n_2}$. Then suppose without loss of generality that $n_1 > n_2$. Then by construction of our original sequence $\{f_n\}$ this means that $||f_{n_1} - f_{n_2}||\geq \epsilon$, i.e. that both $f_{n_1}$ and $f_{n_2}$ cannot lie in the same ball of radius $\epsilon/2$ about $\varphi$. So this explains why $B_{\epsilon/2}(\varphi)$ can only contain at most one term of your subsequence, say this term is $f_{n_k}$

But then if we now choose $B_{||\varphi -f_{n_k}||}(\varphi)$ then this will contain no point of our subsequence, contradicting the fact that $\{f_{n_j}\}$ converges to $\varphi$.

But then we have produced an example of a sequence living in $\mathcal{H}$ that has no convergent subsequence hence contradicting our hypothesis. It follows that $\mathcal{H}$ is totally bounded.

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