# Thread: Subsequences and Subsequential Limits

1. ## Subsequences and Subsequential Limits

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' = $\displaystyle \varnothing$, then S' $\displaystyle \subseteq$ S, so S is closed. Otherwise,
Given any x $\displaystyle \in$ S', where N*(x,$\displaystyle \epsilon$) represents the deleted neighborhood of x of length $\displaystyle \epsilon$:
For any $\displaystyle \epsilon$ > 0, N*(x,$\displaystyle \epsilon$/2) $\displaystyle \cap$ S $\displaystyle \neq$ $\displaystyle \varnothing$, so there is some y in the preceding intersection.
y $\displaystyle \in$ S, so for some subsequence tn of sn, tn converges to y. Then for some M, n > M $\displaystyle \rightarrow$ |tn - y| < $\displaystyle \epsilon$/2.
We know that 0 < |y - x| < $\displaystyle \epsilon$/2, so 0 < |y - x| + |tn - y| < $\displaystyle \epsilon$.
Then, by triangle inequality, |tn - x| < $\displaystyle \epsilon$

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 $\displaystyle \neq$ y.

My goal here was to show that within any $\displaystyle \epsilon$ x, there's some other subsequential limit y within $\displaystyle \epsilon$/2 of x since x is an accumulation point (limit point), and within $\displaystyle \epsilon/2$ of y, there has to be some element of some subsequence of sn. Thus, within any $\displaystyle \epsilon$ of x, we can always find some element of sn that is $\displaystyle \epsilon$ away from x, so x must be a subsequential limit and therefore in S.

Any help in finding my error is greatly appreciated!

2. ## Re: Subsequences and Subsequential Limits

Originally Posted by Lord Voldemort
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' = $\displaystyle \varnothing$, then S' $\displaystyle \subseteq$ S, so S is closed. Otherwise,
Given any x $\displaystyle \in$ S', where N*(x,$\displaystyle \epsilon$) represents the deleted neighborhood of x of length $\displaystyle \epsilon$:
For any $\displaystyle \epsilon$ > 0, N*(x,$\displaystyle \epsilon$/2) $\displaystyle \cap$ S $\displaystyle \neq$ $\displaystyle \varnothing$, so there is some y in the preceding intersection.
y $\displaystyle \in$ S, so for some subsequence tn of sn, tn converges to y. Then for some M, n > M $\displaystyle \rightarrow$ |tn - y| < $\displaystyle \epsilon$/2.
We know that 0 < |y - x| < $\displaystyle \epsilon$/2, so 0 < |y - x| + |tn - y| < $\displaystyle \epsilon$.
Then, by triangle inequality, |tn - x| < $\displaystyle \epsilon$

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 $\displaystyle \neq$ y.
I don't think that you have an error in that work.
It well known that is a Hausdorff space that the set of accumulation points of a set is closed.
The only possible difficulty that I see is showing that if $\displaystyle t$ is an accumulation point of S then $\displaystyle t$ is subsequential limit of $\displaystyle s_n$.