Urysohn's Lemma

The idea is to define a sequence of open sets with the closure of the previous set contained in the next set of the sequence, as we know that it is possible to uniquely extend continuous a function on a set \(A\) to the set \(\overline{A}\). In particular, we should index this sequence by the rationals, which are countable, ensuring that

\[p < q \Longrightarrow \overline{U_p} \subset U_q\]

Defining the sequence inductively is clearly the path that must be taken. We let \(P\) denote the set of rationals in \([0, 1]\), as Munkres does, with \(p_0 = 1\) and \(p_1 = 0\). We let \(U_1 = X - B\), an open set, and from normality, we choose \(U_0\) to be an open set containing \(A\) which is disjoint from a neighbourhood of \(B\), and thus has \(\overline{U_0} \subset U_1\). Thus, we have defined \(U_p\) satisfying the desired condition for the first two elements of \(P\).

Assume we have define \(U_p\) up until \(U_{p_n}\), we wish to define the case of \(U_{p_{n + 1}}\). Let \(P_n\) denote all rationals \(\{p_0, p_1, \dots, p_n\}\). Clearly, we can separate \(P_n\) into sets \(P_L\) and \(P_R\), where \(P_R\) contains the rationals less than \(p_{n + 1}\), and \(P_R\) contains the ones greater than \(p_{n + 1}\). Clearly, \(P_L\) has a greatest element, while \(P_R\) has a least element. Let these elements be \(U_L\) and \(U_R\) respectively. Note that \(\overline{U_L} \subset U_R\). We choose \(U\) to be an open set containing \(\overline{U_L}\) disjoint from a neighbourhood of \(X - U_R\), so \(\overline{U} \subset U_R\). Thus, \(U_{p_{n + 1}} = U\) satisfies the desired condition, and we have constructed our sequence.

Moving along, we define \(U_p\) for all rationals \(p \in \mathbb{Q}\). This is easy: we let \(U_p = \emptyset\) for \(p < 0\) and \(U_p = X\) for \(p > 1\).

Now, we have all the pieces needed to construct our continuous function. We define the function \(\mathbb{Q}(x)\) as follows. Let

\[\mathbb{Q}(x) = \{ p \ | \ x \in U_p\}.\]

Such a set is lower-bounded by \(0\), as \(x \notin U_p\) for any \(p < 0\), since in this case, \(U_p = \emptyset\). Thus, it has a greast lower-bound, and \(f(x) = \inf \mathbb{Q}(x)\) is well-defined. We claim that this is the continuous function on \(X\) that we need. Indeed, note that for \(x \in A\), \(x\) is contained in \(U_0\), so \(f(x) = 0\). Similarly, for \(x \in B\), we have \(x \in U_1\) and not in any \(U_p\) for \(p \in [0, 1)\), so \(f(x) = 1\).

Now, we must demonstrate that \(f\) is in fact continuous. Let \((a, b) \subset [0, 1]\). Suppose \(f(x) \in (a, b)\). By definition, note that \(\inf \{p \ | \ x \in U_p\} = f(x)\), so for any rational \(r > f(x)\), there exists rational \(r' < r\) such that \(x \in U_{r'}\). Since \(U_{r'} \subset U_r\), it follows that \(x \in U_r\) for all \(r > f(x)\). In addition, it is clear that \(f(x) \notin U_r\) for \(r < f(x)\), or else \(\inf \{p \ | \ x \in U_p\}\) would be less than \(f(x)\). We choose rational \(a'\) and \(a''\) such that \(a < a' < a'' < f(x)\). Since \(\overline{U_{a'}} \subset U_{a''}\), and \(f(x) \notin U_{a''}\), \(f(x) \notin \overline{U_{a'}}\), so \(f(x) \in X - \overline{U_{a'}}\), which is open.

Let \(U = (X - \overline{U_{a'}}) \cap U_{b'}\), where \(b'\) is rational with \(f(x) < b' < b\). Clearly, \(U\) is open and \(x \in U\). Moreover, note that given \(y \in U\), we must have \(\inf \{p \ | \ y \in U_p\} \leq b'\), and similarly, it must be greater than \(a'\), so \(f(y) \in (a, b)\). Thus, by definition, \(f\) is continuous.