A point in a proof about sequential compactness

[radou]

OK, so I'm going through the proof of a theorem in Munkres:

Theorem 28.2. If X is a metrizable space, then the following statements are equivalent:

(1) X is compact
(2) X is limit point compact
(3) X is sequentially compact

There's an argument in (2) ==> (3) which I don't quite get. I'll quote directly:

Assume X is limit point compact. Given a sequence (xn) of points in X, consider the set A = {xn : n is a positive integer}. If the set is finite, then there is a point x such that x = xn, for infinitely many values of n. In this case, the sequence (xn) has a constant subsequence, and therefore converges trivially.

OK, if the set A is finite, we have a finite sequence. By definition, a subsequence yn of a given sequence is obtained from the original one by taking an increasing sequence of positive integers n1 < n2 < n3 < ... and defining yi = xni.

I don't understand how we obtained a constant infinite sequence from a finite sequence?

Thanks for any help on this one, it's really bugging me.

 

[micromass]

No, I think you interpreted that wrong. A sequence ALWAYS has infinitely many elements. But if A is finite, then this sequence can only assume finitely many elements.

Take the sequence $$x_n=(-1)^n$$. This has infinitely many elements (which coincide). But the sequences image is finite.

In the proof we have that A is finite. So the sequence (although having an infinity of elements) has a finite image. So there has to be inifitely many elements of the sequence which are sent to the same element. So we have found an infinite subsequence which is constant.

I wish I could explain this better...

 

[tjackson3]

What micromass said is correct. What you have to understand is that there are infinitely many elements of a sequence; however, if that sequence is taken as a set, then the set may be finite. I'll steal his example. Consider the sequence $x_n = (-1)^n$. That sequence looks like -1,1,-1,1,... for all natural numbers n, and since there are infinitely many natural numbers, the sequence is infinite. However, as a set, the sequence is $\{-1,1\}$. So there exists some $x \in x_n$ such that $x = x_n$ for infinitely many values of n. Note that it doesn't say for all n, or for all n > some N, but for infinitely many. In this example, if you let $x = 1$, then $x = x_2,x_4,... = x_{2n}$. Clearly there are infinitely many such instances of this occurring.

This is a subsequence, and since it is constant, it converges. If there is any way to form a constant subsequence out of a sequence, then it converges, as is the case for a periodic sequence like this.

n.b.: Obviously a sequence taken as a set is not always finite. Consider $a_n = \frac{1}{n}$. As a set, this is $\{1,1/2,1/3,...\}$.

 

[radou]

Gee, this was really simple. Thanks a lot :)