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A point $p$ in a topological space $X$ is said to be generic if $\overline{\{p\}} =X$ (i.e. $\{p\}$ is dense in $X$).

Let $G(X)=\{p \mid p\text{ is generic in }X\}$. That is, $G(X)$ is the set of all the points dense in $X$.

$X$ path-connected if $G(X)$ is nonempty.

Show $G(X)$ is a compact subspace of $X$.

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Yes, it was part (a) of the question which I had to prove. (Show X is path-connected if G(X) is nonempty) –  Matt Dec 6 '11 at 15:33
    
Not sure if that's even relevant to part b but put it in there in case! –  Matt Dec 6 '11 at 15:34
    
Hint: "$\{ p \}$ is dense" means "every nonempty open set contains $p$". –  Chris Eagle Dec 6 '11 at 15:36
    
Yes, p is an element of every open set of X was also used for part (a). I tried to work with it but dislike where I was going! –  Matt Dec 6 '11 at 15:39
    
If G(x) is finite is easy, I am running into a problem if G(X) is infinite –  Matt Dec 6 '11 at 15:41

1 Answer 1

There isn't much to add to Chris' hint in the comment. Since every nonempty open set contains every point of $G(X)$, given any cover of $G(X)$, every element of the cover contains all of $G(X)$, so every element forms a finite subcover of $G(X)$ by itself.

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Why does every nonempty open set contain EVERY point of G(X)? –  Matt Dec 6 '11 at 15:55
    
Nevermind, I understand! Every point in G(X) is an element of each open set, which implies every open set contains every element of G(X)! –  Matt Dec 6 '11 at 16:01
    
I was reading too much into it! Thanks guys! –  Matt Dec 6 '11 at 16:02

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