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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)2011-12-06
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    Not sure if that's even relevant to part b but put it in there in case!2011-12-06
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    Hint: "$\{ p \}$ is dense" means "every nonempty open set contains $p$".2011-12-06
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    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!2011-12-06
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    If G(x) is finite is easy, I am running into a problem if G(X) is infinite2011-12-06
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    Ahh I got turned around. You're quite right about (a). How bizarre!2011-12-06
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    If G(X) is finite I consider the smallest open sets containing each p in G(X) which implies a finite subcollection of open covers. But what if G(X) is infinite?2011-12-06

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