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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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It's hard to determine what exactly are you asking: is it that you can't grasp the proof in Rudin's book, or else that you "imagine" the theorem shouldn't, or ought not, be true? –  DonAntonio Nov 16 '12 at 3:26
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In $\mathbb R$ (or $\mathbb R^n$) you can give a bisection-style proof that I think is pretty intuitive. –  littleO Nov 16 '12 at 3:27
    
@DonAntonio I mean I understand the proof I think on a very superficial level. It follows from previous theorems. We look at the range of $\{p_n\}$, which is a subset of a compact space. If it is finite, the proof is pretty obvious. If it is infinite, then it has a limit point somewhere in $X$. Maybe this is what I'm having trouble with—maybe I need to go back to that theorem. But what it is is just that I can't imagine how such a subsequence would be constructed out of the type of sequence that I've described. –  crf Nov 16 '12 at 3:31
    
@littleO Can you elaborate on that? –  crf Nov 16 '12 at 3:31
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I wrote an answer that goes into more detail. –  littleO Nov 16 '12 at 3:47
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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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Wow, this really helps! Just one question though—does this depend on the compactness of our disk? What if it was an open disk? Couldn't we make the same argument? –  crf Nov 16 '12 at 3:57
    
@crf: Dear crf, If the disk were open (but of finite radius), then the limiting value of $r$ could be the outer radius, and so then our subsequence would converge to a point that wasn't in our open disk (but was on its boundary). (The point is that the closure of our open disk, which is just the corresponding closed disk, is compact.) If our open disk were infinite radius (i.e. the whole plane), then the points could escape off to infinity, and so have no convergent subsequence at all. (Think about how the "bisection argument" would fail if you tried to apply it to the interval ... –  Matt E Nov 16 '12 at 4:00
    
... $[0,\infty)$.) Regards, –  Matt E Nov 16 '12 at 4:01
    
Ahh, of course! I forgot exactly what we were trying to prove, hah. This is perfect, thank you so much for your detailed reply. It's funny that there's "not enough room" for a sequence which doesn't converge to a point in $X$ when $X$ is a closed ball, but when you "peel" an infinitely thin layer off of $X$, suddenly there is. –  crf Nov 16 '12 at 4:03
    
@crf: Dear crf, You're welcome, and I'm glad it helped, and that you enjoyed thinking it through. Cheers, –  Matt E Nov 16 '12 at 4:04
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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.) –  littleO Nov 16 '12 at 3:52
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