Show that every infinite closed subset of \(\mathbb R^2\) is the closure of a countable set.
Solution
Let \(F\subseteq\RR^2\) be an infinite closed set.
1. The family\[ \mathcal B = \{B(q,r):q\in\QQ^2,\ r\in\QQ_{>0}\} \]is a countable basis for the Euclidean topology on \(\RR^2\).
The sets \(\QQ^2\) and \(\QQ_{>0}\) are countable, so \(\mathcal B\) is countable. Given an open set \(U\subseteq\RR^2\) and \(x\in U\), choose \(\varepsilon>0\) with \(B(x,\varepsilon)\subseteq U\). Density of \(\QQ^2\) and \(\QQ\) permits a rational center \(q\) sufficiently close to \(x\) and a positive rational radius \(r\) such that \[ x\in B(q,r)\subseteq B(x,\varepsilon). \] Thus \(\mathcal B\) is a basis.
2. There is a countable set \(C\subseteq F\) meeting every\(B\in\mathcal B\) for which \(B\cap F\neq\varnothing\).
For each basis element \(B\in\mathcal B\) satisfying \(B\cap F\neq\varnothing\), choose one point \[ x_B\in B\cap F, \] and let \(C\) be the set of all chosen points. Since \(\mathcal B\) is countable by step 1, so is \(C\). By construction, \(C\subseteq F\).
3. The set \(C\) is dense in \(F\).
Fix \(x\in F\) and an open neighborhood \(U\) of \(x\) in \(\RR^2\). By step 1, there is \(B\in\mathcal B\) with \[ x\in B\subseteq U. \] Since \(x\in F\), the intersection \(B\cap F\) is nonempty. Step 2 therefore provides \(x_B\in C\cap B\subseteq C\cap U\). Thus every neighborhood of \(x\) meets \(C\), so \(x\in\overline C\).
4.\[ \overline C=F. \]
Step 3 gives \(F\subseteq\overline C\). Conversely, step 2 gives \(C\subseteq F\), and \(F\) is closed, so \(\overline C\subseteq F\).
5. Q.E.D.
The set \(C\) is countable by step 2 and has closure \(F\) by step 4.