E is infinite subset of compact set, then is E' also a subset?

Question

Here's a theorem in Rudin's Principles of Mathematical Analysis.

2.37 Theorem: If E is an infinite subset of a compact set K, then E has a limit point in K.

Proof: If no point of K were a limit point of E , then each q∈K would have a neighborhood Vq which contains at most one point of E. It is clear that no finite subcollection of Vq can cover E; and the same is true of K, since E⊂K. This contradicts the compactness of K.

I understood the proof, but My question is this: Is it also true for the statement If E is an infinite subset of a compact set K, then E' (the set of limit points of E ) is subset of K ??

Thm 2.37 only says that there exists a element of E' which is in K, so I made a conjecture that statement is false, but it is hard to think of the counterexample. Is it true or not? and why it is?

Answer 1

Yes, $E' \subseteq K$. This holds because $K$, being compact (in a metric space) is a closed set.

And a set $A$ is closed iff $A' \subseteq A$, which Rudin also proves in the book.

So in your case: $E' \subseteq K' \subseteq K$, where the first follows from $E \subseteq K$, and the second from $K$ being closed.

Answer 2

If the space is Hausdorff, then compact sets are closed and then the answer is yes. Simply because it means that the closure of $E$ is a subset of the compact set.

However, once you realize this, it becomes very easy to find a counterexample. For example, take any space with more than one point which has a dense singleton (e.g., the reals with the topology that a nonempty set is open if and only if $0$ is an element of the set), then your singleton is compact, but has limit points outside the set.

Now you can turn this example to a space where there is an infinite set which is both compact and dense, but not the entire space.