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Let $X$ have a countable basis; let $A$ be an uncountable subset of $X$. Show that uncountably many points of $A$ are limit points of $A$.

Although I find this question asked on MSE, I believe my proof is sufficiently different; and I believe it is correct, but I suppose you'll be the judge. I am, however, having trouble justifying one step, which I will point out at the end of my proof:

Suppose that $X$ has a countable basis $\mathcal{B}$ (i.e., that it is second countable, and let $A$ be some uncountable subset of $X$. Then $X$ must also be first countable which means that every point $x$ has a countable basis $\mathcal{B}_x$. Let $x \in A-E$ and $\mathcal{B}_x$ its associated basis. Note that $\bigcup_{x \in A} \mathcal{B}_x \subseteq \mathcal{B}$ is a uncountable union of countable sets contained in a countable set which is a contradiction, unless there exists a $z \in A$ such that $\mathcal{B}_y = \mathcal{B}_z$ for uncountably many $y \in A$. Letting $E$ denote the collection of such $y$, it's clear that $E$ consists of some of $A$'s limit points: for if $y \in E$, and $B \in \mathcal{B}_y$ is arbitrary, then $z \in B$ and therefore $B \cap (A - \{y\})$ is not empty.

In the MSE chatroom, Ted Shifrin informed me that means that an uncountable union of countable sets being contained in a countable set implies at least one of the sets in the union occurs uncountably many times, which I took to mean there exists a $\mathcal{B}_x$ such that $\mathcal{B}_y = \mathcal{B}_x$ for uncountably many $y \in A$. How does one prove this?

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  • $\begingroup$ there is a logical problem here... it's not completely clear whether $x$ is fixed or not, in fact there might exists $x$ such that $\mathcal{B}_x\not = \mathcal{B}_y$ for every $y\not = x$. All you can prove is that there exists an uncountable set $\Omega$ in $A$ such that $\mathcal{B}_x=\mathcal{B}_y$ whenever $x,y\in \Omega$, and this is enough for your arguments. $\endgroup$ – Yanko Oct 25 '17 at 19:15
  • $\begingroup$ @yanko I understand the confusion. I used $x$ in two different ways. In the first instance, I used it to set up the notation I used in that uncountable union; in the second instance, it's a constant. I should have used to different symbols. Let me fix that. $\endgroup$ – user193319 Oct 25 '17 at 19:18
  • $\begingroup$ Ted Shifrin was wrong, if you transcribed his statement accurately. The uncountable union of all subsets of the naturals is a countable set, but each of them occurs just once in the union. $\endgroup$ – user21820 Oct 26 '17 at 14:24
  • $\begingroup$ Okay I checked the transcript and you transcribed his statement wrongly. He was answering to your specific question and never made a general claim. $\endgroup$ – user21820 Oct 26 '17 at 14:49
  • $\begingroup$ Re: "Although I find this question asked on MSE". Could perhaps add a link? Other users might be interested in solutions in other posts about the same question. (As a side not, if the main purpose of your post is checking your own proof - as opposed to asking for any proof of this claim - you should use (proof-verification) tag.) $\endgroup$ – Martin Sleziak Oct 26 '17 at 14:59
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It cannot be the case that $\mathcal{B}_x = \mathcal{B}_y$ for $x \neq y$ in a $T_0$ space: in that case some $O$ exists such that $x \in O, y \notin O$ (or the other way round) and no element of $\mathcal{B}_y$ can be contained in $O$ but some element of $\mathcal{B}_x$ must be, so $\mathcal{B}_x \neq \mathcal{B}_y$.

Your approach should be changed.

Let $N$ be the set of points of $A$ that are not limit points of $A$ (the uncountable set). So for each $x \in N$ we have some open set $O$ such that $x \in O$ and $O \cap A = \{x\}$, and we can pick a base element $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq O$ and so $B_x \cap A = \{x\}$. For $x \neq y$ this implies $B_x \neq B_y$. So the map $x \to B_x$ is 1-1 from $N$ to $\mathcal{B}$ and the latter is a countable set, so $N$ is at most countable. It follows that $A \setminus N$ is the uncountable subset of $A$ that are limit points.

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  • $\begingroup$ I think his mistake is that he do not know that $\mathcal{B}_x\subseteq \mathcal{B}$. $\endgroup$ – Yanko Oct 25 '17 at 19:21
  • $\begingroup$ I don't believe I am assuming $X$ is a $T_0$ space; honestly, I am not exactly sure what that is. $\endgroup$ – user193319 Oct 25 '17 at 19:21
  • $\begingroup$ @yanko all $\mathcal{B}_x$ are just the members from $\mathcal{B}$ that contain $x$. $\endgroup$ – Henno Brandsma Oct 25 '17 at 19:21
  • $\begingroup$ @user193319 I just mean that you cannot show in general that $\mathcal{B}_x =\mathcal{B}_y$ at all, it will fail in almost all spaces. $\endgroup$ – Henno Brandsma Oct 25 '17 at 19:22
  • $\begingroup$ Ah, fine I understand... he can't prove that $\mathcal{B}_x=\mathcal{B}_y$, he can only prove that one of the open sets $U$ in one of the $\mathcal{B}_x$ appears uncountably many times right? $\endgroup$ – Yanko Oct 25 '17 at 19:26
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Following Henno's answer I have an explicit counter-example.

Take $\mathbb{R}$ with the standard topology, and $\mathcal{B}$ be all the intervals $(q,p)$ for $q,p\in\mathbb{Q}$, this is a countable basis.

For every $x$, $\mathcal{B}_x$ is the set of intervals $(q,p)$ for $q<x<p$. Which is countable.

Take $A=(0,1)$, then for every $x\in A$ you have that $\mathcal{B}_x\subseteq \mathcal{B}$ (obviously). And $\bigcup_{x\in A} \mathcal{B}_x \subseteq \mathcal{B}$, but not $\mathcal{B}_x = \mathcal{B}_y$ because if $x\not = y$ then you can take $q,p$ sufficiently close to $x$ such that $(q,p)$ contains $x$ but doesn't contain $y$, in particular $(q,p)\in \mathcal{B}_y$ but not in $\mathcal{B}_x$.

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