Convergence of a sequence in a compact metric space We know that every infinite subset of a compact metric space has a limit point in that metric space. 
Let $X$ be a compact metric space. A sequence $\{p_n\}$ can be thought of as an infinite subset of $X$. Thus it should have a limit point in $X$ and hence must be convergent.
Is this true or have I made any mistake?
 A: This is false. It is true that a sequence $(p_n)$ can be associated to the set of points $\{p_n\}$ in the sequence, and provided that $(p_n)$ does not repeat too often this set is infinite. (One example how this would fail is if $X = [0,1]$ and $p_n = 0$ if $n$ is even, $p_n = 1$ if $n$ is odd. Then the associated set of points is $\{0,1\}$.)
Even if the associated set of points is infinite, all you can conclude is that this set has a limit point in the space, not that the sequence converges. For example, let $p_n = 1$ when $n$ is even, $p_n = \frac{1}{n}$ when $n$ is odd. Then the associated set of points in the sequence is $\{1\}\cup\{\frac{1}{n}: n~\text{is odd}\}$. This set has $0$ as a limit point. However, the associated sequence does not converge, though it does have a convergent subsequence.
In fact, in a metric space limit point compactness is equivalent to sequential compactness, so you would be able to conclude that every sequence in a compact set has a convergent subsequence whose limit is contained in the set. Without further information this would be the best that you can conclude.
A: Having a limit point does not necessarily mean it is convergent. It does mean that there is a subsequence that is convergent though.
So every infinite sequence has a convergent subsequence. Note that this is the definition of "sequentially compact", which is equivalent to "compact" in the metric space setting.
