Let $X$ be a compact metric space. Show that every sequence has a convergent subsequence.

Proof (revised version):

Let $(x_n)_{n\in\mathbb{N}}\subseteq X$ be a sequence. Let $\varepsilon >0$ be arbitrary. For every $\varepsilon$ is then $\bigcup_{x\in X} B_\varepsilon (x)\supseteq X$ an open cover.

Since $X$ is compact there exists a finite subcover of $n_\varepsilon$ open balls with centerpoint $y^\varepsilon_1,\dotso, y^\varepsilon_{n_\varepsilon}$

$\bigcup_{i=1}^{n_\varepsilon} B_{\varepsilon}(y^\varepsilon_i)\supseteq X$.

Hence for every $\varepsilon$ at least one $B_\varepsilon(y^\varepsilon_i)$ consists of infinite elements of $(x_n)$ and thus $(x_n)$ has a convergent subsequence.

Is this proof done sufficiently? Thanks in advance.

  • $\begingroup$ Nope. It looks like you're abusing notation. I don't know why the centerpoints of the finite subcover belong to the original sequence. And I also don't see how "for every $\epsilon$ at least one $B_\epsilon(x_i)$ consists of infinite elements of $X$" implies "$(x_n)$ has a convergent subsequence". $\endgroup$ Aug 24, 2019 at 1:31
  • 3
    $\begingroup$ Obviously it is not sufficient since you have never actually obtained a convergence subsequence. Your last sentence just states the conclusion out of nowhere. $\endgroup$ Aug 24, 2019 at 1:35
  • $\begingroup$ This is not sufficient; first you should change $B_{\epsilon}(x_i)$ to $B_{\epsilon}(y_i)$. (The centers of the balls need not be in the sequence, as @mathworker21 points out.) You have infinitely many points within $\epsilon$ of some $y_i$, but the points in that ball do not necessarily form a convergent subsequence. What if $\epsilon$ changes? Then you get a new ball and a new set of points in it. $\endgroup$
    – Dzoooks
    Aug 24, 2019 at 1:37
  • $\begingroup$ Thanks for your comments. But the idea of taking such an open cover for every $\varepsilon$ should be correct? $\endgroup$
    – Cornman
    Aug 24, 2019 at 1:40
  • $\begingroup$ math.stackexchange.com/questions/431146/… $\endgroup$ Jul 24, 2022 at 11:03

4 Answers 4


You seem to have one of the right core idea with using pigeonhole principle, but there are serious issues with the details.

You started with the family of all $\varepsilon$-balls, and you know from compactness that you can find a finite subfamily of them that cover the space. You decided to denote the centres of this finite family by $x_1, \ldots, x_{n_\varepsilon}$, which is not permissible, since $x_1, \ldots x_{n_{\varepsilon}}$ already refers to the first $n_\varepsilon$ terms of the sequence $(x_n)$, which you fixed at the beginning of the proof. You have no reason to assume that the centres of your balls lie on the sequence points.

As I said, the pigeonhole principle idea is the right idea. If $y_1, \ldots, y_n$ are centres of open $\varepsilon$-balls that cover the space, then infinitely many terms of the sequence (as opposed to $X$, as your proof states) must enter one of the balls. This will lead to the right proof.

Then your proof goes straight to the conclusion. There is an issue with this too. You've established an infinite set of sequence points that are all within $2\varepsilon$ distance from each other. But, this doesn't necessarily imply the existence of a convergent subsequence. As it stands, you're defining a different subsequence for each $\varepsilon$, when you really need a single subsequence that works for each $\varepsilon$.

Consider, instead of using arbitrary $\varepsilon$, using $\varepsilon = \frac{1}{m}$, where $m \in \Bbb{N}$. Define your subsequence recursively. Start with $\varepsilon = \frac{1}{1}$, and by the pigeonhole principle, there must be a ball $B_1$ of radius $1$ containing infinitely many points of the sequence $(x_n)$. Also let, $$S_1 = \{k \ge 1 : x_k \in B_1\}.$$ Note that $S_1$ is infinite, hence non-empty, and has a minimum element. We can make this minimum element $n_1$, so that $x_{n_1}$ is the first term in the subsequence.

Then, since $S_1$ is infinite, another pigeonhole principle application applied to it (as opposed to our original sequence) will yield another ball $B_2$, now of radius $\frac{1}{2}$, so that $x_k \in B_2$ for infinitely many $k \in S_1$. Let $$S_2 = \{k > n_1 : x_k \in B_2\}$$ and let $n_2$ be the minimum of $S_2$.

Continue this process, and you'll obtain a sequence of nested sets $$S_1 \supseteq S_2 \supseteq \ldots,$$ and their minimum elements form a subsequence $x_{n_1}, x_{n_2}, \ldots$ (if you've done it right, you should be able to justify why $n_1 < n_2 < n_3 < \ldots$, which is necessary to call it a subsequence).

Why is this subsequence convergent? And what does it converge to? Well, answering the latter question first, the closure of the sequence of open balls $(\overline{B}_m)$ form a nested sequence of compact sets, which must have a non-empty intersection. Since their diameters (which are less than or equal to $\frac{2}{m}$) converge to $0$, this intersection must be a singleton. Let $x$ be the unique point in this intersection.

Why is the subsequence converge to $x$? Well, for every $\varepsilon > 0$, find some $m$ such that $\frac{2}{m} < \varepsilon$. Then, $$k \ge m \implies x_{n_k} \in \overline{B}_m \implies d(x, x_{n_k}) \le \frac{2}{m} < \varepsilon,$$ since $x \in \overline{B}_m$ (it's in all the $\overline{B}_i$s).

As you can see, there are some serious steps missing. The core ideas are pigeonhole principle, and Cantor's intersection theorem (guaranteeing the point of convergence).

  • $\begingroup$ Thank you. I understand the flaws in my proof now, and I am a little bit surprised that the proof of this theorem turned out that technical, but it makes sense to do it that way. $\endgroup$
    – Cornman
    Aug 24, 2019 at 2:09
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    $\begingroup$ Why does $S_1$ have a minimum element when the sequence is {1/n}? $\endgroup$ Aug 24, 2019 at 2:15
  • $\begingroup$ @WilliamElliot What do you mean? Are you asking for me to follow through the first step of the construction using a concrete example such as $\{1/n\}$ in, say, $[0, 1]$? $\endgroup$ Aug 24, 2019 at 2:17
  • $\begingroup$ @WilliamElliot (If it helps, recall that $S_1$ is a subset of $\Bbb{N}$, as it contains only indices of the sequence, not points in the sequence.) $\endgroup$ Aug 24, 2019 at 2:20
  • $\begingroup$ @TheoBendit. Ok, that avoids use of the axiom of choice. $\endgroup$ Aug 24, 2019 at 4:41

Your proof is not sufficient, Theo's proof is much better. To approach it from the angle of a general topologist: Let $(x_n)$ be a sequence in $X$. If $A=\{x_n: n \in \Bbb N\}$ is finite, we have a constant (hence convergent) subsequence by the pigeonhole principle.

So we can assume that $A$ is countably infinite and compactness gives us a point of full accumulation, i.e. some $x \in X$ such that for every open neighbourhood $O$ of $x$ we have that $A \cap O$ is also infinite. (If such an $x$ would not exist, for each $x \in X$ we'd have an open neighbourhood $U_x$ with $U_x \cap A$ finite; now take a finite subcover etc. ....)

Now take open balls $O=B(x,\frac{1}{k})$ for succesive $k$ and pick (by a simple recursion) an increasing sequence of indices $n_k$ such that $x_{n_k} \in B(x, \frac{1}{k})$ which gives us a convergent subsequence converging to $x$.

(Pick $n_1$ such that $x_{n_1} \in A \cap B(x,1)$. Having chosen $n_1 < n_2 < \ldots n_k$ already, consider $B(x, \frac{1}{k+1})$ and we know it has infinite intersection with $A$, so there infinitely many indices $m$ with $x_m \in B(x,\frac{1}{k+1})$. Let $n_{k+1}$ be the first such index larger than $n_k$ and the recursion continues.)


It may be easier to argue by contradiction, as follows: if $X$ is compact metric and $\{x_n\}_n$ has no convergent subsequence, then it has no limit point in $X$, hence is closed. Also, for each integer $n$, there is an open ball $B(x_n)$ such that $B(x_n)\cap \{x_n\}=\{x_n\}.$ It follows that the cover $\{X\setminus \{x_1, x_2,\cdots \}, B(x_1),B(x_2),\cdots\}$ has no finite subcover.


Here is yet another way to prove this.

Define a family $\{ U_k \}_{k \in \mathbb{N}}$ of open sets by $U_k := X \setminus \overline{ \{ x_n \}_{n \geq k} } $ . Notice that by definition $U_k \subseteq U_{k+1}$ and thus for any finite subset $J \subseteq \mathbb{N}$ $$ \bigcup_{j \in J} U_j = U_{\max J} \neq X $$ Since $X$ is compact, this implies $\bigcup_{k \in \mathbb{N} } U_k \neq X$. So let $x \in X$ be some point which is not in the before mentioned union. If $x$ appears infinitely often in $(x_{n})_{n \in \mathbb{N}}$, the constant sequence $(x)_{n \in \mathbb{N}}$ is a convergent subsequence. Otherwise we know that \begin{align} \forall k \in \mathbb{N}: ~~~~~x \notin U_k &~\Longrightarrow~ x \in \overline{ \{ x_n \}_{n \geq k} } \\&~\Longrightarrow ~\forall \varepsilon >0 : ~~B(x,\varepsilon) \cap \{ x_n \}_{n \geq k} \neq \emptyset \end{align} Starting with $\mu(1) := 1$ we recursively define $\mu(k)$ to be the smallest number $n$ with $$ x_n \in B\Big(x,\frac{d(x,x_{\mu(k-1)})}{k} \Big) \cap \{ x_n \}_{n \geq k} \neq \emptyset ~~~~\text{and}~~~~ x_n \neq x $$ By construction, $(x_{\mu(k)})_{k \in \mathbb{N}}$ is then a subsequence which converges to $x$.


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