Check the details of the proof of Theorem 34.2 (the imbedding theorem for a space admitting a countable family of continuous functions that separates points from closed sets).
Solution
Let \(\{f_n:X\to\mathbb R\}_{n\ge1}\) be a countable family of continuous functions that separates points from closed sets, and define \[ F:X\longrightarrow\mathbb R^\omega, \qquad F(x)=(f_1(x),f_2(x),\ldots), \] where \(\mathbb R^\omega\) has the product topology. We verify that \(F\) is an embedding.
First, \(F\) is continuous. A subbasic open set in the product has the form \[ \pi_n^{-1}(U), \] with \(U\subset\mathbb R\) open, and \[ F^{-1}(\pi_n^{-1}(U))=f_n^{-1}(U), \] which is open in \(X\).
Second, \(F\) is injective. If \(x\ne y\), the singleton \(\{y\}\) is closed under the hypotheses of the theorem (the space is regular, hence \(T_1\)). Since the family separates points from closed sets, some \(f_n\) satisfies \[ f_n(x)\notin \overline{f_n(\{y\})}=\{f_n(y)\}. \] Thus \(F(x)\ne F(y)\).
It remains to prove that \(F^{-1}:F(X)\to X\) is continuous. It is enough to show that for every open \(U\subset X\) and every \(x\in U\), there is a product-open set \(W\subset\mathbb R^\omega\) such that \[ F(x)\in W \quad\text{and}\quad F^{-1}(W)\subset U. \] Let \(C=X-U\), a closed set not containing \(x\). By separation, choose \(f_n\) such that \[ f_n(x)\notin\overline{f_n(C)}. \] There is therefore an open interval \(V\subset\mathbb R\) containing \(f_n(x)\) and disjoint from \(f_n(C)\). Set \[ W=\pi_n^{-1}(V). \] Then \(W\) is product-open, contains \(F(x)\), and if \(z\in F^{-1}(W)\), then \(f_n(z)\in V\), so \(z\notin C\), hence \(z\in U\). Thus \[ F(U)=F(X)\cap \bigcup_{x\in U}W_x \] is open in \(F(X)\). Therefore \(F\) is a homeomorphism from \(X\) onto the subspace \(F(X)\subset\mathbb R^\omega\), as required.