If I have a compact set $A$ and a closed subset $\Sigma \subset A$ which only contains isolated points (that is, none of them is a limit point). Does the compactness of $A$ then force $\Sigma$ to have finite cardinality ?
Here is my attempt at a proof that the above question can be answered in the positive:
Suppose for contradiction that $\Sigma$ contains infinitely many distinct points.
EDIT :
Then we can construct a sequence of points in $\Sigma$ which consists of distinct points.
By compactness of A, this sequence must have a convergent subsequence, and by the fact that $\Sigma$ is closed, this limit lies in $\Sigma$. But then it cannot be a limit point, because all points in $\Sigma$ are isolated. So the subsequence must eventually constant and equal to the limit, contrary to the construction of the sequence.
Is the reasoning above correct ? If no, what did go wrong ?
Best Answer
What you did is not quite correct even assuming that you're working in a first countable space, so that compactness is equivalent to sequential compactness. Assuming that $\Sigma$ is infinite to get a contradiction is fine (if unnecessary), but that doesn't make $\Sigma$ a sequence: it has no ordering, and it might even be uncountable. However, if $\Sigma$ is infinite there must be a sequence $\langle x_n:n\in\Bbb N\rangle$ of distinct points of $\Sigma$. Then this sequence must have a convergent subsequence, and your argument goes through from there. It's a small point, but it's a good idea to get into the habit of precision in order to avoid confusing yourself when things get more complicated.