Prove or disprove: the set \(\mathbb Q\) of rational numbers is the intersection of a countable family of open subsets of \(\mathbb R\).
Solution
1. Suppose, toward a contradiction, that there are open sets\[ U_1,U_2,\ldots\subseteq\RR \]such that\[ \QQ=\bigcap_{n=1}^{\infty}U_n. \]
This is the negation of the conclusion to be proved.
2. Every \(U_n\) is dense in \(\RR\).
By step 1, \[ \QQ\subseteq U_n \] for every \(n\). The rational numbers are dense in \(\RR\), so every set containing \(\QQ\) is dense.
3. Enumerate the rationals as\[ \QQ=\{q_1,q_2,\ldots\} \]and define\[ V_n=U_n\sm\{q_n\}. \]Then every \(V_n\) is open and dense in \(\RR\).
The set \[ \RR\sm\{q_n\} \] is open and dense. Hence \[ V_n = U_n\cap(\RR\sm\{q_n\}) \] is open.
By step 2, \(U_n\) is open and dense, and \(\RR\sm\{q_n\}\) is also open and dense. The intersection of two open dense subsets of \(\RR\) is dense, so \(V_n\) is dense.
4. One has\[ \bigcap_{n=1}^{\infty}V_n=\varnothing. \]
Let \(x\in\RR\).
If \(x\) is irrational, then by step 1, \[ x\notin\bigcap_{n=1}^{\infty}U_n, \] so \(x\notin U_m\) for some \(m\), and hence \(x\notin V_m\).
If \(x\) is rational, then \(x=q_k\) for some \(k\). By definition, \[ q_k\notin V_k. \]
Thus no real number belongs to every \(V_n\).
5. Step 4 contradicts the Baire category theorem.
The real line \(\RR\) is a complete metric space. By the Baire category theorem, a countable intersection of open dense subsets of \(\RR\) is dense, and in particular nonempty. Step 3 says that every \(V_n\) is open and dense, while step 4 says their intersection is empty. This is impossible.
6. Therefore the statement is false:\[ \boxed{ \QQ\text{ is not a countable intersection of open subsets of }\RR. } \]
The assumption in step 1 led to the contradiction in step 5.
7. Q.E.D.
Step 6 gives the required disproof.