One of my book problems is:
Let sn be a bounded sequence and let S denote the set of subsequential limits of sn. Prove that S is closed.
Here's my attempt:
Let S' be the set of accumulation points (limit points) of S.
If S' = , then S' S, so S is closed. Otherwise,
Given any x S', where N*(x, ) represents the deleted neighborhood of x of length :
For any > 0, N*(x, /2) S , so there is some y in the preceding intersection.
y S, so for some subsequence tn of sn, tn converges to y. Then for some M, n > M |tn - y| < /2.
We know that 0 < |y - x| < /2, so 0 < |y - x| + |tn - y| < .
Then, by triangle inequality, |tn - x| <
What I don't understand here is that this seems to show that tn converges to x which can't be true because tn converges to y by definition, and x y.
My goal here was to show that within any x, there's some other subsequential limit y within /2 of x since x is an accumulation point (limit point), and within of y, there has to be some element of some subsequence of sn. Thus, within any of x, we can always find some element of sn that is away from x, so x must be a subsequential limit and therefore in S.
Any help in finding my error is greatly appreciated!