Talk:Alexander's Compactness Theorem

PM: you reverted my edit to the version that didn't make any sense. Why'd you do that? --Dfeuer (talk) 08:24, 11 February 2013 (UTC)

Small formal improvement
Hi,

although I agree that I understood the proof as is, it might be a small improvement to change the line

That means that if $V \notin \mathcal C$, then $\mathcal C \cup V$ has a finite subcover, necessarily of the form $\mathcal C_0 \cup V$.

to

That means that if $V \notin \mathcal C$, then $\mathcal C \cup \{V\}$ has a finite subcover, necessarily of the form $\mathcal C_0 \cup \{V\}$ for some finite subfamily $\mathcal C_0$ of $\mathcal C$.

What do you think? --Dondon (talk) 06:16, 18 September 2016 (EDT)


 * Yes, good call. It also occurs to me that $V$ is not defined, and it needs to be. Not sure whether it's an open set, I think it is. Please feel free to fix this up. Cheers. --prime mover (talk) 06:26, 18 September 2016 (EDT)