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.