Closed subset of compact set

Question

Consider the following statement: Any closed subset of a compact set is compact. A standard way to prove this statement goes like this:

Suppose $K$ is a compact set in the metric space $E$. Let $S\subseteq K$ be any closed set. Let $(U_i)_{i\in I}$ be any open cover of $S$. Then $(U_i)_{i\in I}$ and $S^c$ forms an open cover of $K$. It has a finite subcover since $K$ is compact. Then $S$ can be covered by finitely many $U_i$s.

I'm a bit confused by the proof. In particular I don't know if "closed" and "open" in the proof are relative to $K$ or $E$. I assume that the statement is saying that the subset $S$ is closed relative to $K$. Then $S^c$ is open relative to $K$. It's not necessarily open in $E$. Why can we conclude that $(U_i)_{i\in I} \cup \{S^c\}$ is an open cover of $K$?

Another question would be is compactness a relative concept (That is, if $S\subseteq K$ is compact in the entire metric space $E$, is it compact in the subspace $K$)?

Answer 1

There are a few equivalent definitions to a subset $K\subseteq E$ being compact:

$1$. If $K=\bigcup\limits_{i\in I}U_i$ where $U_i$ are open subsets of $K$ then this cover must have a finite subcover.

$2$. If $K\subseteq\bigcup\limits_{i\in I}U_i$ where $U_i$ are open subsets of $E$ then this cover must have a finite subcover.

The proof that these two conditions are equivalent is straightforward from the definitions, you should try to prove it. In particular, the first condition shows that $K$ being compact depends only on the topology on $K$ as a subspace of $E$. Now, if $S\subseteq K$ then the topologies on $S$ induced from $E$ and from $K$ are the same, and so $S$ being compact in $K$ is equivalent to $S$ being compact in $E$. (and both are simply equivalent to $S$ being compact as a topological space) That answers your second question.

As for the first question, note that in a metric space, compact subsets are closed. So $K$ is closed in $E$, and therefore, $S\subseteq K$ is closed in $K$ if and only if it is closed in $E$. So you can choose how to think of $S$. The important thing is just to be consistent when you continue the proof. If you think of $S$ as being closed in $E$ then you should take the $U_i$ as open subsets of $E$, and the complement $S^c$ is also with respect to $E$. If that is your choice then here you work with the second definition that I gave for compactness.