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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?

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    $\begingroup$ 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. $\endgroup$
    – qualcuno
    Feb 2 at 4:12
  • $\begingroup$ @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. $\endgroup$ Feb 2 at 4:14
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    $\begingroup$ 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 $\endgroup$
    – qualcuno
    Feb 2 at 4:18
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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 $f[A]=\{1\}$.

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  • $\begingroup$ This makes sense to me. Thanks! $\endgroup$ Feb 2 at 4:43
  • $\begingroup$ @EnricoBorba: You’re welcome! $\endgroup$ Feb 2 at 4:50

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