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This is theorem 3.6 in Rudin's Principles of Mathematical Analysis. I actually followed the logical steps of the proof, but I can't really picture why it's true like I can with some of these other theorems. Intuitively I feel like it shouldn't be so hard to construct a counterexample. I'm imagining that $X$ is, say, a closed ball in $\mathbb{R}^2$, and then we have a sequence which follows some weird spirally path all over and around the inside of $X$—think a spirograph where the radius of the inner circle varies randomly or something. I can't imagine how it is that such a sequence would have a convergent subsequence. Is there something wrong with my example? Or am I just not seeing something that I ought to be?

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    Its sort of a recursive pigeon-hole principle.2012-11-16

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It might be a good idea to really write down an example of the kind you describe, and see how it necessarily contains a convergent subsequence.

So let's suppose that our compact subset is the closed disk (ball) in $\mathbb R^2$, which I'll identify with $\mathbb C$ (so that it's easy to use polar coordinates), and let our sequence be $r_n e^{i \theta_n}$. In your example, you imagine that the $r_n$ are varying like crazy, and so are the the $\theta_n$s.

Still, we will find a convergent subsequence!

First, let's focus on the $r_n$. They are a sequence in the interval $[0,1]$. Now the bisection proof that litteO mentions in a comment applies. That is, cut $[0,1]$ into the two intervals $[0,1/2]$ and $[1/2, 1]$. At least one of these contains infinitely many of the $r_n$. Continuing with such bisections, we find a subsequence $r_{n_i}$ that converges to some $r \in [0,1]$. Let's replace our sequence $r_n e^{i \theta_n}$ by the subsequence $r_{i_n} e^{i\theta_{n_i}}$, and then relabel, so that we may assume in our original sequence $r_n e^{i \theta_n}$ that the radii $r_n$ are actually converging to some fixed radius $r$.

Now if the limiting radius if $r = 0$, then in fact our sequence $r_n e^{i\theta_n}$ is converging to $0$, and we're done. If $r \neq 0,$ then what we know is that the points $r_n e^{i \theta_n}$ get closer and closer to the circle of radius $r$, but we don't have any control of their angles $\theta_n$. But now we can focus on the $\theta_n$ (forgetting the $r_n$ for a moment), and making the same bisection argument, on the interval $[0,2\pi]$ now, we can find a subsequence of angles $\theta_{n_i}$ which converge to a limit, say $\theta$, and so passing to this subsequence, we find that it converges to $r e^{i\theta}$.

Conclusion: no matter how crazily $r_n$ and $\theta_n$ vary, the bisection argument shows that some subsequence of the $r_n$ has to tend towards a limit, and then for some subsequence of this subsequence, the angles also have to tend towards a limit (unless the subsequence of $r_n$ tends towards $0$, in which case we don't have to worry about the angles at all).

The bisection argument shows that as long as we are looking at quanities in a bounded closed interval, there is just "not enough space" to avoid the existence of a convergent subsequence, however much the original sequence is varying.

Additional remark: Of course, one can prove the result in the case of a disk just by embedding it into a square, and putting a finer and finer mesh over the square (the two-dimensional version of the bisection argument). I phrased the above argument in terms of the polar coordinates because it seemed to fit with the kind of counterexample you were trying to imagine.

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    @crf: Dear crf, You're welcome, and I'm glad it helped, and that you enjoyed thinking it through. Cheers,2012-11-16
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In $\mathbb R$ (or $\mathbb R^n$) you can give an intuitive bisection-style proof.

Let $I = [a,b] \subset \mathbb R$ be a closed interval and let $(p_n)_{n=1}^{\infty}$ be a sequence of numbers in $I$. Let $L_1 = [a,\frac{a+b}{2}]$ and $R_1 = [\frac{a+b}{2},b]$. If $L_1$ contains infinitely many points of $(p_n)_{n=1}^{\infty}$, then let $I_1 = L_1$. Otherwise, let $I_1 = R_1$. So $I_1$ contains infinitely many points of $(p_n)_{n=1}^{\infty}$.

Now subdivide $I_1$ in the same way, and continue in this manner, creating an infinite nested sequence of closed intervals $I \supset I_1 \supset I_2 \supset \cdots$. The intersection $\cap_{i=1}^{\infty}$ is non-empty and contains a single point $p$. You can see that there is a subsequence of $(p_n)_{n=1}^{\infty}$ which converges to $p$.

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    A technical point: for this argument, I should assume the sequence $(p_n)_{n=1}^{\infty}$ takes on infinitely many values. (If this is not the case, showing existence of a convergent subsequence is easy.)2012-11-16