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Can I get feedback/help with my proof please? Thanks!

Let $A$ be a subset of $\mathbb{R}^n$ for $n\ge 1.$ Let $B(A)$ denote the points of $A$ such that $p\in B(A)$, then there is an open set $U$ with $p\in U$ and $U\cap A$ is countable. Prove $B(A)$ is countable.

$\textit{Proof.}$ Let $p\in B(A)$ be arbitrary. Then $p\in A$ such that there is an open set $U$ with $p\in U$ and $U\cap A$ is countable. So as $p\in A$ and $U\cap A$, $p\in U\cap A$ as $p$ was arbitrary and $p\in A$, $p\in U\cap A$. Therefore, $B(A)\subseteq U\cap A$ as $U\cap A$ is countable. Therefore, $B(A)$ is countable as it is contained in a countable set.

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    $\begingroup$ You can't include $\;B(A)\;$ in the definition of $\;B(A)\;$ ... $\endgroup$
    – DonAntonio
    May 15, 2020 at 23:37

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For each $p\in B(A)$ there is an open set $U_p$ such that $p\in U_p$ and $U_p\cap A$ is countable. It is true that $p\in U_p\cap A$, since $p\in U_p$ and by definition $B(A)\subseteq A$, but that in no way implies that $B(A)\subseteq U_p$. I suspect that you were thinking that you had one open set $U$ that worked for every $p\in B(A)$, but in fact it’s entirely possible that $U_p\ne U_q$ whenever $p,q\in B(A)$ and $p\ne q$.

Big HINT: You will have to use the fact that $\Bbb R^n$ is second countable, i.e., that it has a countable base $\mathscr{B}$. For each $p\in B(A)$ there is a $B_p\in\mathscr{B}$ such that $p\in B_p\subseteq U_p$. Is $B_p\cap A$ countable? How many different basic open sets $B_p$ are there?

Added: Let $\mathscr{B}=\{B_k:k\in\Bbb N\}$ be a countable base for $\Bbb R^n$. For each $p\in B(A)$ there is a $k(p)\in\Bbb N$ such that $p\in B_{k(p)}\subseteq U_p$; $B_{k(p)}\cap A\subseteq U_p\cap A$, so $B_{k(p)}\cap A$ is countable.

Now let $V=\bigcup_{p\in B(A)}B_{k(p)}$; then

$$V\cap A=\left(\bigcup_{p\in B(A)B_{k(p)}}B_{k(p)}\right)\cap A=\bigcup_{p\in B(A)}(B_{k(p)}\cap A)$$

is a union of countable sets. How many countable sets? Each $k(p)$ is a natural number, and there are only countably many natural numbers, so there are only countably many different sets $B_{k(p)}\cap A$. For a given $\ell\in\Bbb N$ there may be many $p\in B(A)$ such that $k(p)=\ell$, but they all have the same $B_{k(p)}\cap A$. Thus, $V\cap A$ is the union of countably many countable sets and as such is countable. And $V\cap A$ contains every point of $B(A)$, so $B(A)$ is countable.

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  • $\begingroup$ It is countable but $\mathbb{R}^n$ is Lindelöf. So theres countable open subcover for every subset of $\mathbb{R}^n$ $\endgroup$
    – brucemcmc
    May 15, 2020 at 23:53
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    $\begingroup$ @brucemcmc: Being Lindelöf is not enough, since $B(A)$ is not known to be closed (and in fact needn’t be). One could use the fact that every second countable space is not just Lindelöf, but hereditarily Lindelöf, but that’s an unnecessary complication and uses a fact that you may not have available at this point; all that you need here is the fact that $\mathscr{B}$ is countable. $\endgroup$ May 15, 2020 at 23:58
  • $\begingroup$ Thanks Brian. I am still confused. Could you please write a proof? $\endgroup$
    – brucemcmc
    May 17, 2020 at 20:17
  • $\begingroup$ @brucemcmc: I’ve now done so. $\endgroup$ May 17, 2020 at 21:03
  • $\begingroup$ Thank you. It is clear to me now $\endgroup$
    – brucemcmc
    May 18, 2020 at 0:26

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