Subsequences and Subsequential Limits

One of my book problems is:

Let s_{n} be a bounded sequence and let S denote the set of subsequential limits of s_{n}. 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 t_{n} of s_{n}, t_{n} converges to y. Then for some M, n > M |t_{n} - y| < /2.

We know that 0 < |y - x| < /2, so 0 < |y - x| + |t_{n} - y| < .

Then, by triangle inequality, |t_{n }- x| <

What I don't understand here is that this seems to show that t_{n} converges to x which can't be true because t_{n} 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 s_{n}. Thus, within any of x, we can always find some element of s_{n} 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!

Re: Subsequences and Subsequential Limits

Quote:

Originally Posted by

**Lord Voldemort** One of my book problems is:

Let s

_{n} be a bounded sequence and let S denote the set of subsequential limits of s

_{n}. 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 t

_{n} of s

_{n}, t

_{n} converges to y. Then for some M, n > M

|t

_{n} - y| <

/2.

We know that 0 < |y - x| <

/2, so 0 < |y - x| + |t

_{n} - y| <

.

Then, by triangle inequality, |t

_{n }- x| <

What I don't understand here is that this seems to show that t

_{n} converges to x which can't be true because t

_{n} converges to y by definition, and x

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 is an accumulation point of S then is subsequential limit of .