Prove that every closed subset \(C\subseteq\mathbb R\) is the closure of a finite or countable set.
Solution
1. Let\[ \mathcal I \coloneqq \{[p,q]:p,q\in\QQ,\ p<q\}. \]Then \(\mathcal I\) is countable.
The set \(\QQ\) is countable, so \(\QQ\times\QQ\) is countable. The family \(\mathcal I\) is indexed by the subset \[ \{(p,q)\in\QQ^2:p<q\}, \] and is therefore countable.
2. For every \(I\in\mathcal I\) with\[ C\cap I\ne\varnothing, \]choose one point\[ x_I\in C\cap I, \]and let\[ D \coloneqq \{x_I:I\in\mathcal I,\ C\cap I\ne\varnothing\}. \]Then \(D\) is finite or countable and\[ D\subseteq C. \]
There is at most one chosen point for each interval in the countable family \(\mathcal I\), so \(D\) is at most countable. Every chosen point belongs to \(C\) by construction.
3. One has\[ \overline D\subseteq C. \]
By step 2, \[ D\subseteq C. \] Since \(C\) is closed, it contains the closure of each of its subsets. Hence \[ \overline D\subseteq C. \]
4. Every point of \(C\) lies in \(\overline D\).
Fix \(x\in C\) and let \(\varepsilon>0\). By density of \(\QQ\) in \(\RR\), choose rational numbers \(p,q\) such that \[ x-\varepsilon<p<x<q<x+\varepsilon. \] Then \[ I=[p,q]\in\mathcal I, \qquad x\in C\cap I, \] so \(C\cap I\) is nonempty and the construction in step 2 supplies \[ x_I\in D\cap[p,q]. \] Because \[ [p,q]\subseteq(x-\varepsilon,x+\varepsilon), \] the \(\varepsilon\)-neighborhood of \(x\) meets \(D\). Since \(\varepsilon>0\) was arbitrary, \[ x\in\overline D. \]
5. Therefore\[ \boxed{\overline D=C}. \]
Step 3 gives \[ \overline D\subseteq C, \] while step 4 gives \[ C\subseteq\overline D. \] Thus equality holds.
6. Q.E.D.
Step 2 shows that \(D\) is finite or countable, and step 5 shows that its closure is \(C\).