Proving $B(A)$ is countable.

Question

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.

Answer 1

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.