# Definition confirmation: cover of a subset

In lectures we defined:

Let $X$ be a topological space, $Y \subset X$ a subset. A collection $\mathcal{A} \subset \mathcal{P}(X)$ is a cover of $Y$ by sets open in X if each element of $\mathcal{A}$ is an open set in $X$ and $Y \subset \cup_{A \in \mathcal{A}}A$.

We then went on to a proposition:

Let $Y$ be a subset of $X$. Then $Y$ is compact if and only if every cover of $Y$ by subsets open in $X$ has a finite subcover.

I think the "subsets" part in the proposition should actually be "sets", but I want to make sure since we could end up with something different otherwise.

(This particular lecturer takes a while to answer emails so I'm asking here instead.)

• What's your objection against subsets? It is somehow redundant if the sets are characterized as "open in X", but quite harmless I think. – drhab Feb 25 '16 at 21:23
• I agree, seems harmless. – sqtrat Feb 25 '16 at 21:25
• @drhab Perhaps I've been thinking too much about this, but it doesn't seem to make strict sense to say "subsets" since we've not defined what it means. – Irregular User Feb 25 '16 at 21:36
• In my view "subsets open in $X$" somehow implicates that we are dealing here with subsets of $X$. According to that interpretation the statement is true. Something else: I would rather use the statement to define compactness of $Y$. – drhab Feb 25 '16 at 21:39

## 1 Answer

Saying that $B$ is a subset of $X$ does not mean that $B$ is not equal to $X$. (We call those proper subsets.) Hence, it does not make any difference whether you call them sets or subsets.