# The Alexandroff compactification is the biggest one-point compactification?

Greets

I don't know whether the following property is true let $X$ be a noncompact topological space and let $\omega$ not belong to $X$. Let $X^*=(X\cup{\omega},\tau^*)$ be the Alexandroff compactification of $X$, then if $\tau^{**}$ is another topology on $X\cup{\omega}$, containing the topology of $X$ with $(X\cup{\omega},\tau^{**})$ compact, then we can embed $(X\cup{\omega},\tau^{**})$ into $(X\cup{\omega},\tau^*)$.

Thanks

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Suppose that $\tau$ is a locally compact, non-compact topology on $X$, $p$ is a point not in $X$, $X^*=X\cup\{p\}$, and $$\tau^*=\tau\cup\big\{X^*\setminus K:K\text{ is compact in }\langle X,\tau\rangle\big\}\;;$$ $\langle X^*,\tau^*\rangle$ is then compact. Suppose further that $\tau'$ is a topology on $X^*$ such that $\langle X,\tau'\rangle$ is compact and $\tau\subseteq\tau'$.

Let $p\in U\in\tau'$, and let $F=X^*\setminus U\subseteq X$. Let $\mathscr{V}$ be any $\tau$-open cover of $F$; then $\mathscr{V}\cup\{U\}$ is a $\tau'$-open cover of $X^*$, so it has a finite subcover $\mathscr{W}$. Clearly $U\in\mathscr{W}$, since no other member of $\mathscr{V}$ contains $p$. Let $\mathscr{W}_0=\mathscr{W}\setminus\{U\}$; then $\mathscr{W}_0$ is a finite subfamily of $\mathscr{V}$ covering $F$, since $U\cap F=\varnothing$. It follows that $F$ is $\tau$-compact and hence that $U\in\tau^*$. Clearly, then, $\tau^*$ and $\tau'$ have the same neighborhoods of $p$. Moreover, $\{U\in\tau^*:U\subseteq X\}=\tau$, and $\{U\in\tau':U\subseteq X\}\supseteq\tau$, so $\tau'\supseteq\tau^*$.

However, it is possible to have $\tau'\supsetneqq\tau^*$. Let $Y=\Bbb N\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the topology $\tau_Y$ that it inherits from the usual topology on $\Bbb R$. (Note that $0\in\Bbb N$.) Let $0'$ be a new point not in $Y$, and let $X=Y\cup\{0'\}$; the topology $\tau$ on $X$ is $$\tau=\tau_Y\cup\big\{(U\setminus\{0\})\cup\{0'\}:0\in U\in\tau_Y\big\}\;.$$ (In other words, $X$ is just $Y$ with the point $0$ split into two points, $0$ and $0'$, with the ‘same’ neighborhoods.) Finally, let $$K=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}\;;$$ note that $K$ is a compact subset of $X$ that isn’t closed in $X$, since $0'$ is in its closure.

Define $p$, $X^*=X\cup\{p\}$, and $\tau^*$ as before, and let $\tau'$ be the topology on $X^*$ generated by $\tau^*$ and $X^*\setminus K$. Then $\langle X^*,\tau'\rangle$ is still compact, since $K$ is compact in $\langle X^*,\tau^*\rangle$, and $\tau'\supsetneqq\tau^*$.

In fact, the example shows that this will happen whenever $\langle X,\tau\rangle$ has a compact subset that isn’t closed.

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I think what you meant was $\left\{{0}\right\}\cup{\left\{{\displaystyle\frac{1}{n}:n\in{\mathbb{Z}^+}}\rig‌​ht\}}$, correcting this your proof, is correct Thank you very much!!!!!!!! –  Camilo Arosemena Oct 23 '12 at 21:34
@Camilo: I did indeed; thanks for catching it. –  Brian M. Scott Oct 23 '12 at 21:39
and also think that what you wanted to show in the first part was that $p$ has the same neighborhoods in $\left<{X,\tau^´}\right>$ and $\left<{X,\tau^*}\right>$; note that not necessarly we have $\tau^*\subseteq{\tau^´}$, because of your counterexamnple; note that the space of your counterexample is locally compact. –  Camilo Arosemena Oct 23 '12 at 21:47
@Camilo: My statement that $\tau^*$ and $\tau'$ have the same nbhds of $p$ means exactly the same as your version. And yes, we do necessarily have $\tau^*\subseteq\tau'$; what my example shows is that we don’t necessarily have $\tau'\subseteq\tau^*$. –  Brian M. Scott Oct 23 '12 at 21:57
sorry, I hadn't seen your "moreover" –  Camilo Arosemena Oct 23 '12 at 23:25