Relatively Compact definition.

Question

Let $A \subset X$, then $A$ is said to be relatively compact if the closure $cl(A)$ is compact in $X$. Why is this called relatively here when it is "relative" to the universal space $X$?

Shouldn't relative compactness mean $A \subset E \subset X$ is relatively compact if $A = E \cap C$ for $C$ compact in $X$.

Answer 1

A set of the form $C \cap E$ where $C$ is compact in $X$ is relatively compact in $E$ (when $X$ is Hausdorff), as $\operatorname{cl}(C \cap E) \subseteq C$ and so is a closed subset of a compact set, hence compact.

If $\operatorname{cl}(A)$ is compact, then we cannot always find such a $C$ with $E \cap C= A$, as that would make $A$ closed in $E$ (again for Hausdorff spaces), and non-relatively-closed sets can also be relatively compact.

So defining relatively compact as you suggests creates a weaker, less useful notion.

Answer 2

Being relatively compact is not an absolute property of a space $A$, but of an embedding $A \subset X$ into an ambient space $X$. It would be more precise to say that $A$ is relatively compact in $X$. If $A$ is relatively compact in $X$ and $A \subset Y$ for some other space, it need not be relatively compact in $Y$ (take e.g. an non-compact $A$ and $Y = A$).

In your definition a third space $E$ enters and there are two interpretations.

1) $A$ is relatively compact in $X$ if there exist $E$ and $C$ as above.

2) $A$ is relatively compact in the a pair $(X,E)$ if there exists $C$ as above.

The first interpretation is equivalent to

3) There exists a compact $C \subset X$ such that $A \subset C$.

1) $\Rightarrow$ 3) is trivial, for the converse take $E = A$. Thus, if $X$ is Hausdorff, then $\text{cl}(A)$ must be compact which is the "usual" definition. More generally, the same is true in any $KC$-space $X$ (see https://topospaces.subwiki.org/wiki/KC-space).

The second interpretation makes things complicated by working with a pair $(X,E)$ instead of a space $X$. There are two "extremal" choices for $E$:

a) $E = X$. This means that $A$ is compact.

b) $E = A$. We get nothing else than 3).

a) generalizes to any closed $E \subset X$. In this case $A$ is a closed subset of $C$, hence itself compact.

For any choice of $E$, 2) implies 3). Thus, for a $KC$-space $X$, 2) implies that $\text{cl}(A)$ is compact. The converse may be true for some choices of $(X,E)$, but a) shows that is in general false.