Closed set in $\mathbb R^n$ is closure of some countable subset

Question

Let A be a closed set in R^n . How can I show that A = closure of B where B is countable ? Thanks for any help .

Answer 1

For $p\in\Bbb R^n$ and $r>0$ let $B(p,r)=\{x\in\Bbb R^n:d(p,x)<r\}$, where $d(p,x)$ is the usual Euclidean distance in between $p$ and $x$ in $\Bbb R^n$. Let $\mathscr{B}=\{B(p,r):p\in\Bbb Q^n\text{ and }r\in\Bbb Q^+\}$, where $\Bbb Q^+$ is the set of positive rationals. $\mathscr{B}$ is a countable family of open sets in $\Bbb R^n$. Let $\mathscr{B}_0=\{B\in\mathscr{B}:B\cap A\ne\varnothing\}$.

For each $B\in\mathscr{B}_0$ choose a point $x_B\in B\cap A$, and let $D=\{x_B:B\in\mathscr{B}_0\}$; clearly $D$ is a countable subset of $A$. I leave it to you to prove that $\operatorname{cl}D=A$; it’s not hard, but I’ll add more explanation if you need it.

(There is in general no way to specify exactly how the points $x_B$ are chosen, so in general this argument requires the axiom of choice.)

Added: To show that $A\subseteq\operatorname{cl}D$, let $x$ be any point of $A$. For any $\epsilon>0$ there are an $r\in\Bbb Q^+$ and a $p\in\Bbb Q^n$ such that $r<\epsilon/2$ and $p\in B(x,r)$. Then $B(p,r)\in\mathscr{B}$, and $x\in A\cap B(p,r)\ne\varnothing$, so $B(p,r)\in\mathscr{B}_0$, and $x_{B(p,r)}\in D\cap B(p,r)$. Finally, $d(x_{B(p,r)},x)\le d(x_{B(p,r)},p)+d(p,x)<2r<\epsilon$, so $x_{B(p,r)}\in D\cap B(x,\epsilon$. Since $\epsilon>0$ was arbitrary, this shows that $x\in\operatorname{cl}D$, and since $x\in A$ was arbitrary, it shows that $A\subseteq\operatorname{cl}D$.

Answer 2

The space $\mathbb R^n$ has a countable base, i.e. it is a second countable space. E.g., the balls with rational diameters and centers having rational coordinates can be taken as a such base.

If $A=\emptyset$, you can take $B=\emptyset$. So let us assume from now on that $A$ is non-empty.

If you take any countable basis $$\mathcal B=\{B_n; n\in\mathbb N\}.$$

Now for each $n$ you take $b_n$ as an arbitrary element of $A\cap B_n$, if $A\cap B_n$ is non-empty.

If $A\cap B_n=\emptyset$, you can choose $b_n$ as an arbitrary element of $A$.

The set $B=\{b_n; n\in\mathbb N\}$ fulfills the condition $\overline B=A$.

• Since $B\subseteq A$ and $A$ is closed, we have $\overline B\subseteq A$.
• If an element $x$ belongs to $A$ then for every basic neighborhood $B_n\ni x$ of $x$ we have $B_n\cap A\ne\emptyset$. This implies that $b_n\in B_n$. Therefore every neighborhood of $x$ contains an element of the set $B=\{b_n; n\in\mathbb N\}$; which means that $x\in\overline B$.

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Notice, that we have used Axiom of Choice since we have chosen one element $b_n$ from each non-empty $A\cap B_n$.

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The above proof is almost identical to the usual proof of the fact that Second-Countable Space is Separable.

Answer 3

For each of the countably many intervals $[a,b]$ with $a,b\in\mathbb Q^n$ select a single point of $[a,b]\cap A$ (if there is one). Then if $x\in A$ and $\varepsilon>0$, find $a,b\in \mathbb Q$ with $x\in [a,b]\subset B(x,\varepsilon)$ and observe that the point chosen $\in[a,b]\cap A$ (which may be $x$ itself) has distance $<\varepsilon$ from $x$.

The above construction uses the axiom of countable choice.