Compactness in a metric space
Let X = the set of real number sequence such that
Let N, let
Define , is compact?
First, we claim that there exist a subsequence of every sequence in that converges in X.
Suppose that is a sequence in , note that
Note that , so is bounded. By a theorem, we know that there exist a subsequence of such that converges to a point in R, denote by .
Now, consider the component in the subsequence , which is also bounded. Thus there exist a subsequence of such that converges to a point in R, this time denote by .
Note that the component is a subsequence of , thus it also converges to the point .
Continue in the manner, we find a subsequence of , denote by for simplification purpose, such that N.
Now, consider . Thus proves that .
Now, we claim that is a point in .
Assume to the contrary that is not a point in . We have
Note that since it is a sequence in .
But, , in which contradicts our assumption. Therefore we show that , thus proves is compact.
Please check my proof, thank you!