Examine the proof of the Urysohn lemma, and show that for given \(r\),
\[ f^{-1}(r) = \bigcap_{p > r} U_p - \bigcup_{q < r} U_q, \]
where \(p, q\) range over the rationals.
Solution
Goal: Prove that the level set \(f^{-1}(r)\) of the Urysohn function \(f(x) = \inf \{p \in \mathbb{Q} \mid x \in U_p\}\) is given by the difference \(\bigcap_{p > r, p \in \mathbb{Q}} U_p \setminus \bigcup_{q < r, q \in \mathbb{Q}} U_q\).
1. Definition and properties of the Urysohn function \(f\):Let \(\{U_p\}_{p \in \mathbb{Q}}\) be the nested collection of open sets satisfying \(\overline{U}_p \subseteq U_{p'}\) for all rationals \(p < p'\), with \(U_p = \varnothing\) for \(p < 0\) and \(U_p = X\) for \(p > 1\). The function \(f: X \to [0, 1]\) is defined by:\[f(x) = \inf \{p \in \mathbb{Q} \mid x \in U_p\}.\]
2. Characterization of the sublevel set \(\{x \in X \mid f(x) \le r\}\):\(f(x) \le r \iff x \in \bigcap_{p > r, p \in \mathbb{Q}} U_p\).Proof:
2.1. If \(f(x) \le r\), then for any rational \(p > r\), since \(\inf\{t \in \mathbb{Q} \mid x \in U_t\} \le r < p\), there exists a rational \(t < p\) with \(x \in U_t\). Because \(t < p\), \(U_t \subseteq U_p\), so \(x \in U_p\). Thus \(x \in \bigcap_{p > r} U_p\).
2.2. Conversely, if \(x \in \bigcap_{p > r} U_p\), then \(x \in U_p\) for every rational \(p > r\), which means \(f(x) \le p\) for all rational \(p > r\). Taking the infimum over all \(p > r\) yields \(f(x) \le r\).
3. Characterization of the strict sublevel set \(\{x \in X \mid f(x) < r\}\):\(f(x) < r \iff x \in \bigcup_{q < r, q \in \mathbb{Q}} U_q\).Proof:
3.1. If \(f(x) < r\), by definition of the infimum, there exists a rational \(q\) with \(x \in U_q\) such that \(q < r\). Hence \(x \in \bigcup_{q < r} U_q\).
3.2. Conversely, if \(x \in \bigcup_{q < r} U_q\), then \(x \in U_q\) for some rational \(q < r\). Then \(f(x) \le q < r\), so \(f(x) < r\).
4. Evaluation of the level set \(f^{-1}(r)\):Proof:
4.1. For any point \(x \in X\), \(f(x) = r\) if and only if \(f(x) \le r\) and \(f(x) \not< r\).
4.2. By 2 and 3:\[f(x) = r \iff x \in \left( \bigcap_{p > r, p \in \mathbb{Q}} U_p \right) \quad \text{and} \quad x \notin \left( \bigcup_{q < r, q \in \mathbb{Q}} U_q \right).\]
4.3. Set-theoretically, this is equivalent to:\[x \in \bigcap_{p > r, p \in \mathbb{Q}} U_p \setminus \bigcup_{q < r, q \in \mathbb{Q}} U_q.\]
5. Conclusion:\[f^{-1}(r) = \bigcap_{p > r} U_p - \bigcup_{q < r} U_q.\]Q.E.D.