Separating closed sets in Moore plane / Niemytzki plane (Topology)

Question

I spent the last few days trying to solve this exercise with little success, so I really hope someone here might be able to assist:

Denote Moore plane by $M$, the $x$-axis by $R$ and the upper half-plane by $H$ ($M = R \cup H$). Let $A, B \subseteq M$ be closed and disjoint subsets of $M$. Suppose $|A \cap R| < \infty$. Prove that $A$ and $B$ can be separated by disjoint open neighborhoods.

I am aware of the fact that $M$ is not $T_4$, and in the previous exercises I proved the following facts:

• $R$ as a subspace of $M$ has the discrete topology.
• $H$ as a subspace of $M$ is homeomorphic to $H$ as a subspace of $\mathbb{R}^2$ (with the standard topology).
• $H$ is open in $M$.
• The closure in $M$ of each element of the topology's basis is the same as its closure in the Euclidean plane.
• $M$ is a regular Hausdorff space ($T_3$).

I also showed that it suffices to prove the claim for the case where $A \cap R = \varnothing$ and $B = R$ (although other approaches might also work), but I'm having trouble showing that there exists an open neighborhood of $A$ whose closure does not intersect with $R$.

Your help will be much appreciated!

Answer 1

Looking at the reduced case where $A\cap R=\emptyset$ and $B=R$, note that $H$ (as a subspace) is homeomorphic to the standard Euclidean open half-plane and as such is normal, and as $H$ is an open neighbourhood of $A$ we find an open (in Euclidean $H$ (so also open in the Niemytzki plane)) subset $U$ with $A \subseteq U \subseteq \overline{U} \subseteq H$, so the closure of $U$ in $H$ (or the Euclidean plane) misses $R$ (i.e. stays inside $H$).

It's easy to check from the definition of topology of the Niemytzki plane that for $x \in R$ and $B \subseteq H$ we have that $x \in \overline{B}$ (Niemytzki topology) implies $x \in \overline{B}$ (Euclidean topology), and this implies that $\overline{U}$ in the Niemytski topology is a subset of the closure of $U$ in the Euclidean topology (and thus misses $R$). So $U$ and $M\setminus \overline{U}$ separate $A$ (disjoint from $R$) and $B=R$, as you needed.

Answer 2

Here is my solution:

Since $A\cap R=\varnothing$, for each $\left(x,y\right)\in A$ it holds that $y>0$, and therefore $\left(x,y\right)\in B_{\frac{y}{3}}\left(\left(x,y\right)\right)\subseteq H$. Denote $W:=\bigcup_{\left(x,y\right)\in A}B_{\frac{y}{3}}\left(\left(x,y\right)\right)\subseteq H$, so $W$ is open in $H$.

Since $A$ is closed in $H$, it follows from $H$'s normality (as a subspace of a metric space) that there exists an open set $U$ ($U$ is open in $H$, but since $H$ is open in $M$, $U$ is also open in $M$) such that $A\subseteq U\subseteq cl_{H}\left(U\right)\subseteq W$. We claim that $cl_{M}\left(W\right)\subseteq H$, and therefore $cl_{M}\left(U\right)\subseteq cl_{M}\left(W\right)\subseteq H$ (as $U\subseteq W$ implies that $cl_{M}\left(U\right)\subseteq cl_{M}\left(W\right)$).

Assume towards a contradiction that $cl_{M}\left(W\right)\cap R\neq\varnothing,$ so there exists $\left(x,0\right)\in R$ such that every open neighborhood of $\left(x,0\right)$ intersects with $W$. We will show that for each $\varepsilon>0$, $A\cap\left[B_{\varepsilon}\left(\left(x,\varepsilon\right)\right)\cup\left(x,0\right)\right]\neq\varnothing,$ implying that $\left(x,0\right)\in cl_{M}\left(A\right)$, contrary to the fact that $cl_{M}\left(A\right)=A$ (as $A$ is closed in $M$) and $A\cap R=\varnothing.$

Let $\varepsilon>0$, by the definition of a closure, there exists some $\left(x',y'\right)\in W\cap\left[B_{\frac{\varepsilon}{2}}\left(\left(x,\frac{\varepsilon}{2}\right)\right)\cup\left(x,0\right)\right]$, and from $W$'s definition it follows that there exists $\left(x'',y''\right)\in A$ such that $d\left(\left(x',y'\right),\left(x'',y''\right)\right)<\frac{y'}{2}$. Note that since $\left(x',y'\right)\in B_{\frac{\varepsilon}{2}}\left(\left(x,\frac{\varepsilon}{2}\right)\right)$, it holds that $\left(x'-x\right)^{2}+\left(y'-\frac{\varepsilon}{2}\right)^{2}<\left(\frac{\varepsilon}{2}\right)^{2}$ and $y'<\varepsilon$.

Now:

$d\left(\left(x,\varepsilon\right),\left(x'',y''\right)\right)\leq d\left(\left(x,\varepsilon\right),\left(x',y'\right)\right)+d\left(\left(x',y'\right),\left(x'',y''\right)\right)<\sqrt{\left(x'-x\right)^{2}+\left(y'-\varepsilon\right)^{2}}+\frac{y'}{2}=\sqrt{\left(x'-x\right)^{2}+\left(y'-\frac{\varepsilon}{2}\right)^{2}+\frac{3\varepsilon^{2}}{4}-\varepsilon y'}+\frac{y'}{2}<\sqrt{\varepsilon^{2}-\varepsilon y'}+\frac{y'}{2}\stackrel{\left(\star\right)}{\leq}\varepsilon $

$\left(\star\right)$ holds because $\sqrt{\varepsilon^{2}-\varepsilon y'}+\frac{y'}{2}\leq\varepsilon$ iff $\sqrt{\varepsilon^{2}-\varepsilon y'}\leq\varepsilon-\frac{y'}{2}$ iff $\varepsilon^{2}-\varepsilon y'\leq\varepsilon^{2}-\varepsilon y'+\left(\frac{y'}{2}\right)^{2}$iff $0\leq\left(\frac{y'}{2}\right)^{2}$ which is always true.

Hence $\left(x'',y''\right)\in B_{\varepsilon}\left(\left(x,\varepsilon\right)\right)\cup\left(x,0\right)$, which implies that $\left(x,0\right)\in cl_{M}\left(A\right)$, a contradiction (as explained above). It follows that $cl_{M}\left(W\right)\cap R=\varnothing$, which implies that $cl_{M}\left(U\right)\subseteq H$ . Therefore, $U$ and $M\backslash cl_{M}\left(U\right)$ are both open in $M$ and separate $A$ and $R$.

Answer 3

Here is a topological proof assuming $A$ is bounded.

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Recall the "one point normalization" of the Niemytzki plane constructed by @BrianMScott in Moore plane / Niemytzki plane (Topology)

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.

Now if $A\subseteq H$ is closed and bounded, it is also closed in $Y$. Notice $R\cup\{p\}$ is closed in $Y$. Since $Y$ is normal, there are disjoint open sets $U$ and $V$ separating $A$ and $R\cup\{p\}$. Then $U$ and $V\setminus \{p\}$ are disjoint open in $X$, separating $A$ and $R$. We are done.