Since $X$ is a compact metric space, it is sequentially compact so every sequence in $X$ has a subsequence which converges. Now, I want to show that $X^\mathbb{N}$ is sequentially compact, but I'm getting confused because we must show that the sequence of a sequence has a convergent subsequence. I know that we must use the fact that $X$ is sequentially compact. Thanks!
If $X$ is a compact metric space, then the set of all sequences in $X$, $X^\mathbb{N}$ is sequentially compact
2
$\begingroup$
general-topology
compactness
-
1Well, $X^N$ is in fact *compact* (for the reasosonable topology...) – 2011-03-17
-
0This is a direct result of Tychonoff's theorem. – 2011-03-17
-
0Yes I need to prove that it is compact and I thought it would be easier to prove that it is sequentially compact. The space I am trying to prove is sequentially compact (or compact since we proved it is a metric space) is the set of all sequences in X. I don't think it is the product of space like R^n. Any further explanantions would be appreciated. It is supposed to involve some diagonalization argument – 2011-03-17
-
0@Mariano Suárez-Alvarez, @Alex: Please see my answer for a reply to your comments. – 2011-03-17
-
0I just reread your comment here and noticed I'd previously overlooked the sentence "I don't think it is the product of space like R^n". The set of all sequences in a space is (isomorphic to) the product of countably many copies of the space. Just like an element of an $n$-fold product is an $n$-tuple, an element of an $\mathbb{N}$-fold product is an $\mathbb{N}$-tuple, i.e. a sequence. – 2011-03-17