Closure of a set in metric space is compact

Question

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!

Answer 1

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 $y$.

Answer 2

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.

Answer 3

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 $j > N$.