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I'm trying to understand a proof about density of a subset $X$ in its one-point compactification $Y$.

We can do this proof by contradiction, suppose we don't have $\operatorname{cl}(X) = Y$. This implies that $\operatorname{cl}(X) = X$.

Why? Can anyone help me?

Thanks

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Suppose $\operatorname{cl}(X)\not=Y$. We know $X\subseteq \operatorname{cl}(X)$ so we get $\operatorname{cl}(X)=X$ and $\infty \notin \operatorname{cl}(X).$ So by definition of closure, there exists a (wlog, open) neighborhood $U$ of $\infty$ s.t. $U \cap X=\emptyset$. The topology of the extension is defined to be all open subsets of $X$ together with all sets $V$ that contain $\infty$ and such that $X\setminus V$ is closed and compact. $\infty \in U$ so $U$ is of the second kind of open sets, meaning $X \setminus U$ is closed and compact. But remember that $U \cap X=\emptyset$ so $X\setminus U=X$. We get that $X$ is compact, in contradiction to the assumption (if $X$ was compact, we wouldn't have needed the compactification).

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    Y is X plus one point. We know that $X \subseteq cl(X)$, or in other words cl(X) is not smaller than X. Also, of course it's contained in our space, Y (it can't contain "elephant" as element). Mathematically, $cl(X) \subseteq Y$. So what cl(X) can be? We're left with two options: X, or Y (because there no options between them - they differ by a point). So if cl(X) isn't Y, it must be X.2012-10-02
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You’re making it much harder than it really is. $Y=X\cup\{p\}$, where $p$ is the new point. To show that $X$ is dense in $Y$, you need only show that every open neighborhood of $p$ has non-empty intersection with $X$. Go back to the definition of the one-point compactification and see why this is true: what are the open neighborhoods of $p$? Why must each of them have non-empty intersection with $X$? It has to do with the fact that $X$ is not compact.