# Theorem about the Moore Plane

I'm going through Topology Through Inquiry by Francis Su and Michael Starbird, and ran into the following exercise:

3.10.3: In $$\mathbb{H}_{bub}$$, let $$A$$ be a countable set on the $$x$$-axis and let $$z$$ be a point on the $$x$$-axis not in $$A$$. Show that there exist disjoint open sets $$U$$ and $$V$$ such that $$A \subseteq U$$ and $$z \in V$$.

Where $$\mathbb{H}_{bub}$$ is the "Upper Half-Plane with the Sticky Bubble Topology", aka the Moore Plane.

I'm confused why this is true because if we take $$A = \mathbb{Q}$$ and $$z = \pi$$, what possible open sets could satisfy 3.10.3? Any radius chosen for the open set containing $$\pi$$ will be too large to not intersect with a rational arbitrarily close to $$\pi$$ given some other radius.

The exercise also asks whether or not the countability requirement on $$A$$ is necessary, but I feel like this is false with any dense set $$A \in \mathbb{R}$$ where $$A \neq \mathbb{R}$$.

I'd love some guidance on this exercise but above all I'd like to understand why my counterexample is wrong.

Edit: as pointed out in the comments, it's clear to me now why my counterexample is correct. We can pick some $$r_\pi$$ for the tangent open ball containing $$\pi$$, and for any rational $$q$$, we can find a radius that doesn't intersect with that open ball. Now the question is why wouldn't this process work for uncountable sets?

• I may think about this later, as for your counterexample: an open neighbourhood of $\pi$ can be $\pi$ and some tangent open ball in the open upper half plane. It won't intersect the $x$ axis except on $\pi$. Now, for each $q \in \Bbb Q$ you can again find a tangent ball to $q$ that does not intersect the chosen one for $\pi$, shrinking the radius enough. Feb 2 at 4:12
• @guidoar Oh jeez... I'm not sure why I was restricting my self to that order of radius selection. For every rational there is definitely a radius where it does not intersect any radius picked for the open set containing $\pi$. Ok this makes sense, thanks for that. I'll think more on the countability as well. Feb 2 at 4:14
• I think we can do the same for any $A$, but I am not entirely sure. Let's call a "tangent ball neighbourhood" of $x$ an open set that contains $x$ and a tangent open ball contained in the upper half plane. If for every $a,b$ in the $x$ axis one can construct disjont open ball neighbourhoods $U_a \ni a, U_b\ni b$ then $V = U_z$ and $U = \bigcup_{a \in A}U_a$ should work. But I haven't thought deeply about it, maybe there's something I am missing Feb 2 at 4:18

Let $$A$$ be any subset of the $$x$$-axis, and let $$z$$ be any point of the $$x$$-axis not in $$A$$. You can indeed separate $$A$$ and $$z$$ by disjoint open sets by first choosing a basic tangent disk nbhd of $$z$$ and then choosing for each $$x\in A$$ a sufficiently small tangent disk nbhd; there is no need to restrict $$A$$ to countable sets.
The result also follows from the fact that the space is Tikhonov: every subset of the $$x$$-axis is closed, so in particular $$A$$ is closed, and there is a continuous $$f:\Bbb H_{\text{bub}}\to[0,1]$$ such that $$f(z)=0$$ and $$f[A]=\{1\}$$.