0
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ What's your objection against subsets? It is somehow redundant if the sets are characterized as "open in X", but quite harmless I think. $\endgroup$ – drhab Feb 25 '16 at 21:23
  • 1
    $\begingroup$ I agree, seems harmless. $\endgroup$ – sqtrat Feb 25 '16 at 21:25
  • $\begingroup$ @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. $\endgroup$ – Irregular User Feb 25 '16 at 21:36
  • 1
    $\begingroup$ 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 $Y$. $\endgroup$ – drhab Feb 25 '16 at 21:39
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.