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I am supposed to show that by adding a point to the Moore plane, that the subspace of a $T_4$-space is not necessarily a $T_4$-space. I do know that the Moore plane is NOT a $T_4-$space. Does anybody here know which point could be meant?

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    I take it that your definition of T$_4$ does not imply the space is Hausdorff? – user642796 Apr 27 '14 at 05:44
  • @ArthurFischer for us: every $T_4$ is $T_2$. (I don't know what one point compactification is.) –  Apr 27 '14 at 15:16
  • @Lipschitz: The one-point compactification won't really help, since the Moore plane is not locally compact the resulting space will not be Hausdorff. – user642796 Apr 27 '14 at 19:16
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    Denoting the Moore plane by $\mathbb{L}$, the idea would be to add some new point, call it $\star$ to $\mathbb{L}$, $X = \mathbb{L} \cup { \star }$, and then topologise $X$ in a manner so that (1) $X$ is compact, and the Moore plane is a subspace of $X$. Basically, you need to ensure that if $E , F \subseteq \mathbb{L}$ are disjoint closed subsets of $\mathbb{L}$ which cannot be separated by open sets, then $\mathrm{cl}X (E) , \mathrm{cl}_X (F)$ are _not disjoint, meaning that $\star$ is an element of both of these closures. (cont...) – user642796 Apr 27 '14 at 19:21
  • (...inued) One idea would be to topologise $X$ so that the open sets are exactly the open subsets of $\mathbb{L}$ as well as $X$. Then $\star$ is an element of every nonempty closed set, and so $X$ satisfies the "normality" condition: any two disjoint closed sets can be separated by disjoint open neighbourhoods. But this space is not Hausdorff. It's not even T$_1$. – user642796 Apr 27 '14 at 19:23

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Let $X$ be the Niemytzki plane, let $H$ be the open upper half-plane, and let $L$ be the $x$-axis. For each $x\in L$ and $r>0$ let $D(x,r)$ be the closed disk of radius $r$ in the usual topology tangent to $L$ at $x$, and let $\mathscr{D}$ be the set of all such disks. Let $p$ be a point not in $X$, and let $Y=X\cup\{p\}$. Topologize $Y$ be making $X$ an open subset with the Niemytzki topology and taking

$$\left\{Y\setminus\bigcup\mathscr{F}:\mathscr{F}\subseteq\mathscr{D}\text{ is finite}\right\}$$

as a local base at $p$. If $U=X\setminus\bigcup_{k=1}^nD(x_k,r_k)$, where $D(x_1,r_1),\ldots,D(x_n,r_n)\in\mathscr{D}$, is a basic open nbhd of $p$, $s_k>r_k$ for $k=1,\ldots,n$, and $V=X\setminus\bigcup_{k=1}^nD(x_k,s_k)$, then $V$ is also an open nbhd of $p$, and $\operatorname{cl}_YV\subseteq U$. It’s clear that $Y$ is $T_1$, and since $X$ is regular, it follows that $Y$ is $T_3$. $Y$ is also Lindelöf: once $p$ is covered by an open set, only finitely many points of $L$ remain to be covered, and $H$ is Lindelöf. It follows at once that $Y$ is $T_4$.

The idea behind this topology is that counterexamples to normality in $X$ are pairs of infinite subsets of $L$; this topology makes $p$ a limit point of every infinite subset of $L$, thereby killing off all such examples.

Brian M. Scott
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