Sequentially compact proof

Hi all.

I am having trouble proving one direction of the following claim:

"Let K be a non-empty subset of R^n. K is sequentially compact
iff it is closed and bounded."

I can prove the "if" direction. I am having trouble with the "only
if". So suppose K is sequentially compact, I want to prove that
K must be closed and bounded.

I thought I'd try contradiction. So suppose K is not bounded,
then for every positive M \in R we can find an element, x_m, in
K such that |x_m| > M. We can construct a sequence, {x_m},
this way. Since {x_m} lies in K, and K is sequentially
compact, {x_m} must have a converging subsequence in K.

Now, since {x_m} is "monotone" and does not converge,
it seems intuitively obvious that no subsequence of {x_m}
can converge. However, I am having trouble turning this
into something more rigorous.

Can anyone help?

Cheers,
Willem


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