Proof that a compact metric space is sequentially compact

I have tried to construct a proof that is different from the others given on this site. It would be great if you could tell me whether the reasoning is sound.

Let $\langle x_{n}\rangle$ be a non-convergent sequence of points. Let the set $\{1,a,b,c,d,\dots\}\in\Bbb{N}$ be strictly increasing.

For $x_{1}$, there is an $\epsilon_{1}\in\Bbb{R}$ such that $B(x_{1},\epsilon_{1})$ does not contain all the points in $\langle x_{n}\rangle\setminus x_{1}$. This is because if for all real numbers $r$ the ball $B(x_{1},r)$ contained every other element in the sequence, then the sequence would be convergent.

Let $x_{a}$ be a point in $\langle x_{n}\rangle\setminus B(x_{1},\epsilon_{1})$. Then there is an $\epsilon_{2}\in\Bbb{R}$ such that $B(x_{i},\epsilon_{2})$ does not contain all the points $x_{j}$ where $j>a$. More importanty, we can ensure that $\epsilon_{2}<\epsilon_{1}$

Let $x_{b}$ be a point in $\langle x_{n}\rangle\setminus B(x_{a},\epsilon_{2})$. Proceeding like this, we can find a non-finite cover of $X$ which does not have a finite subcover, as each set contains a point ($x_{i},i\in\{1,a,b,c,\dots\}$) that no other does.

Is this argument sound?

Thanks for your time!

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I think you have to choose $\langle x_n \rangle$ not only non-convergent, but also such that no subsequence of $\langle x_n \rangle$ converges in $X$. Further I unfortunately don't really see with which sets you cover $X$. – Nils Matthes Jul 11 '13 at 10:33
In a compact space any infinite set has an accumulation point. – egreg Jul 11 '13 at 10:40

2 Answers

The basic idea of your argument is ok, except that (as pointed out by Ittay) you are not producing a cover. However, if you add the complement of the sequence $(x_n)$ to your cover, then you are fine. Note that this is an open set because the sequence has no convergent subsequence by assumption.

A somewhat simpler argument would be to consider the open sets

$$U_k := X\setminus\{x_j:j\geq k\}.$$

These form a cover of your space. Since the sequence must be infinite, none of these sets equals the whole space, and hence there is no finite subcover.

Of course this boils down to the same idea as any other proof.

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Briliant! Thanks – Ayush Khaitan Jul 11 '13 at 12:30

Firstly, instead of considering the set $\{1,a,b,c,d,\cdots \}$, which then forces you to define $x_a,x_b$ etc. just consider the set $\mathbb N$, and then you simply define $x_2,x_3$ and so on.

Regardless, it is not clear what is the cover of $X$ that your argument produces.

I must add that I don't know which other proofs you are referring to, but the proof I have in mind is very short.

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@IttayWeiss- I am referring to this proof. I have not studied completeness yet, if that is what you had in mind. Thanks! – Ayush Khaitan Jul 11 '13 at 10:32
yes, notice that the answer gives a very short proof that a compact metric space is sequentially compact. The rest of the proof shows the converse. You were not asking about the converse. – Ittay Weiss Jul 11 '13 at 10:40
haha true! Should've noticed that. I guess the question now is how I should find a cover for my space. Working on that. – Ayush Khaitan Jul 11 '13 at 10:46
No. My proof is erroneous. $B(x_{c},\epsilon_{3})$ may contain $x_{1}$. In fact we may have a finite cover of $\langle x_{n}\rangle$. – Ayush Khaitan Jul 11 '13 at 11:04
that is not the reason why your proof is wrong (you can choose $\epsilon_3$ small enough.... You proof is incorrect since it does not necessarily produce a covering of $X$. – Ittay Weiss Jul 11 '13 at 11:05