# Closure of a set in metric space is compact

I'm trying to prove that if $X$ is a metric space, then for a set $A\subset X$, the closure $\bar A$ is compact iff for every sequence $(x_n)_{n=1}^\infty\subset A$ there exists a subsequence $(x_{n_k})_{k=1}^\infty$ that converges to some $x\in X$.

The forward direction is pretty straightforward as a particular case of the definition of compactness by B-W property.

For the reverse, let $a\equiv(x_n)_{n=1}^\infty\subset \bar A$ be some sequence. Then, if $a$ has any subsequence in $A$ we are done by assumption. Otherwise, there are at most finitely many points in $a$ that are in $A$. So there exists some subsequence which resides entirely in $\bar A \backslash A$, and also for some $N\in \mathbb N$ the $N$-tail of $a$ is in $\bar A\backslash A$.

But I don't see where this gets me. What guarantees that I can come up with a convergent subsequence? The closure of a set contains all its limit points, but why shouldn't I be able to choose a non-convergent sequence this subset? Or in $\bar A$ in general?

Other approaches such as finite open covers or being a closed subset of a compact set don't seem to be relevant here.

Thanks!

• You might get some mileage out of a diagonal trick: each $x_n\in\bar A$ has its own sequence $\{x_{n,i}\}\subset A$ that converges to $x_n$. Now consider $\{x_{n,n}\}$.... Commented Mar 29, 2017 at 21:28
• @Greg Martin Could you please explain a little further? I didn't quite get it. Specifically, who's ${x_n,n}$? Do you mean that If I have a sequence, then each point is a limit point of some other sequence, so I can pick only points from the original sequence that coincide with the corresponding point in the ${x_n}$ sequence? I'm not sure I am guaranteed to get a non-empty set out of it, nor a convergent one.
– Yoni
Commented Mar 30, 2017 at 7:07

PART 1. Let $\bar A$ be compact and let $(x_n)_{n\geq 0}$ be a sequence in $A.$

Let $C_0$ be a finite cover of $\bar A$ by open balls of $X$ of radius $1.$ Let $\beta_0\in C_0$ such that $\{n:x_n\in \beta_0\}$ is infinite. Let $\gamma_0=\bar {\beta_0}\cap \bar A$ .

Now recursively suppose that $\gamma_m$ is a compact subset of $\bar A$ and that $\{n:x_n\in \gamma_m\}$ is infinite. Let $C_{m+1}$ be a finite cover of $\gamma_m$ by open balls of $X$ of radius $2^{-(m+1)}$ and let $\beta_{m+1}\in C_{m+1}$ such that $\{n:x_n\in \beta_{m+1}\}$ is infinite. And let $\gamma_{m+1}=\bar \beta_{m+1}\cap \gamma_m.$

Now we may take a strictly increasing $f:\mathbb N_0\to \mathbb N_0$ such that $x_{f(n)}\in \gamma_n$ for every $n.$ Now $d(x_{f(n)},x_{f(n')})<2^{-n}$ when $n<n'$, so the sequence $(x_{f(n)})_{n\geq 0}$ is a Cauchy sequence in the closed set $\gamma_0$ , so it converges to a limit point $x$.

PART 2.(Converse). Let $\bar A$ be non-compact. Let $C$ be an open cover of $\bar A$ with no finite subcover.

CASE 1. $C$ has no countable subcover.

For each $p\in \bar A$ take $q_p\in \mathbb Q^+$ and $c_p\in C$ such that $B(p,q_p)\subset c_p.$ For $q\in \mathbb Q^+$ let $$S(q)=\{p\in \bar A: q_p=q\}.$$ Suppose that for every $q\in \mathbb Q^+$ there is a finite $T(q)\subset S(q)$ such that $\cup_{p\in T(q)}B(p,q)\supset S(q).$ Then $\cup_{q\in \mathbb Q^+}\{c_p:p\in T(q)\}$ is a countable sub-cover of $C$, contrary to hypothesis.

Therefore there exists $q_0\in \mathbb Q^+$ such that no finite subset of $\{B(p,q_0):p\in S(q_0)\}$ covers $S(q).$

So take $p_1\in S(q_0)$ and take $p_{n+1}\in S(q_0)$ \ $\cup_{j\leq n}B(p_j,q_0).$ We now have $d(p_n,p_m)\geq q_0$ whenever $n\ne m$.

Finally take $x_n\in A$ such that $d(x_n,p_n)<q_0/4.$ Then $(x_n)_n$ is a sequence in $A$ with $d(x_n,x_m)\geq q_0/2$ whenever $n\ne m$, so it has no convergent subsequence.

CASE 2. $C$ has a countable subcover. Let $\{c_n:n\in \mathbb N\}$ be a countable subcover of $C.$ Since $C$ has no finite subcover, let $$x_n\in \bar A \ \cup_{j<n}c_j.$$ Suppose $(x_n)_n$ had a subsequence $(x_{f(n)})_n$ converging to $x$ (with $f :\mathbb N\to \mathbb N$ strictly increasing).

We have $x\in \bar A$ so $x\in c_{n_0}$ for some $n_0.$

But then $B(x,r)\subset c_{n_0}$ for some $r>0,$ and for all but finitely many $n$ we have $x_{f(n)}\in B(x,r)$ (because $(x_{f(n)})_n$ converges to $x$).... So there exists $n$ such that $f(n)>n_0$ and $x_{f(n)}\in c_{n_0}$, contrary to $x_{f(n)}\not \in \cup_{j<f(n)}c_j.$

Therefore $(x_n)_n$ has no convergent subsequence. Finally let $y_n\in A$ with $d(x_n,y_n)<2^{-n}.$ Then $(y_n)_n$ has no convergent subsequences for if $(y_{f(n)})_n$ converged to $y$ then $(x_{f(n)})_n$ would also converge to $y$.

• Thanks, but isn't this the forward direction? I want to prove $\bar A$ is compact given the what is known about $A$.
– Yoni
Commented Mar 30, 2017 at 7:02
• I added the converse, which can be done more easily using the theory of ordinals and transfinite recursion, but I dk whether you know that stuff, and it's basically the same proof. An equivalent to this result is that a metric space is non-compact iff it has an infinite closed discrete subspace (which is not the case for all topological spaces). Commented Mar 30, 2017 at 9:02
• I'm familiar with ordinals etc., but the above is sufficient! Thank you
– Yoni
Commented Mar 30, 2017 at 9:28

Another approach is the following: let $$(x_n)_n$$ be a sequence in $$\bar{A}$$. Then for every $$n\in\mathbb{N}$$, choose $$y_n\in A$$ such that $$d(x_n,y_n)<1/n$$. Then by hypothesis there is subsequence $$(y_{n_k})_k$$ such that $$y_{n_k}\to y\in \bar{A}$$. Then it's easy to show that $$x_{n_k}\to y$$ also, so $$\bar{A}$$ is compact.

This is a proof using the ideas of @MathNewbie and @GregMartin.

We will use the following theorem (Barry Simon, Real Analysis, Theorem 2.3.6) (Gerald Folland, Real Analysis, comment on Theorem 4.29)

Theorem: Let $$X$$ be a metric space. Then, the following are equivalent:

• $$X$$ is compact
• $$X$$ is sequentially compact (i.e. every sequence has a convergent subsequence)

Then the question can be rephrase as follows.

$$X$$ is a metric space, then for a set $$A\subset X$$, the closure $$\bar A$$ is sequentially compact iff for every sequence $$(x_n)_{n=1}^\infty\subset A$$ there exists a subsequence $$(x_{n_k})_{k=1}^\infty$$ that converges to some $$x\in X$$.

• ($$\Rightarrow$$) Let $$A \subset X$$ and $$\{x_n\}_{n = 1}^{\infty} \subset A$$. Then, $$\{x_n\}_{n = 1}^{\infty} \subset \overline{A}$$ and $$\overline{A}$$ is sequentially compact so there exists a convergent subsequence that converges in $$\overline{A} \subset X$$.

• ($$\Leftarrow$$) Let $$\{x_n\}_{n = 1}^{\infty} \subset \overline{A}$$. Then, for all $$n \in \mathbb{N}$$, since $$X$$ is a metric space and $$x_n \in \overline{A}$$, there exist $$\{y_{nk}\}_{k = 1}^{\infty} \subset A$$ such that $$d(y_{nk}, x_n) < \frac{1}{k}$$. By hypothesis, the sequence $$\{y_{nn}\}_{n = 1}^{\infty}$$ has convergent subsequence, say $$\{y_{n_j n_j}\}_{j = 1}^\infty$$ that converges to $$y_\infty \in \overline{A} \subset X$$.
Let's prove that $$\{x_{n_j}\}_{j = 1}^{\infty}$$ converges to $$y_\infty$$. Indeed, let $$\varepsilon > 0$$. Let $$N_1 \in \mathbb{N}$$ such that $$d(y_{n_j n_j}, y_\infty) < \frac{\varepsilon}{2}$$ for all $$j > N_1$$. Let $$N_2 \in \mathbb{N}$$ such that $$n_j > \left\lceil \frac{1}{\varepsilon} \right\rceil$$ so $$d(y_{n_j n_j}, x_{n_j}) < \frac{\varepsilon}{2}$$ for all $$j > N_2$$. Let $$N = \max\{N_1, N_2\}$$, then $$d(x_{n_j}, y_\infty) \le \underbrace{d(x_{n_j}, y_{n_j n_j})}_{<\frac{\varepsilon}{2}} + \underbrace{d(y_{n_j n_j}, y_\infty)}_{\frac{\varepsilon}{2}} < \varepsilon$$ for all $$j > N$$.