Topologically complete spaces

1.1. Let \(X\) be topologically complete, so there is a metric \(d\) on \(X\) compatible with its topology such that \((X, d)\) is a complete metric space.

1.2. Let \(A \subseteq X\) be a closed subspace, equipped with the subspace metric \(d_A = d|_{A \times A}\).

1.3. Let \((x_n)_{n=1}^\infty\) be a Cauchy sequence in \((A, d_A)\).

1.4. Since \(d_A(x_n, x_m) = d(x_n, x_m)\), \((x_n)\) is a Cauchy sequence in the complete space \((X, d)\).

1.5. Therefore \((x_n)\) converges to some limit point \(x \in X\).

1.6. Since \(A\) is closed in \(X\), every limit of points in \(A\) belongs to \(A\), so \(x \in A\).

1.7. Thus every Cauchy sequence in \((A, d_A)\) converges in \(A\), so \((A, d_A)\) is complete.

2.1. Let \(\{X_n\}_{n=1}^\infty\) be a countable family of topologically complete spaces, with each \((X_n, d_n)\) being a complete metric space.

2.2. Define the bounded metrics \(\bar{d}_n(x, y) = \min\{1, d_n(x, y)\}\) on each \(X_n\).

2.3. On the product space \(X = \prod_{n=1}^\infty X_n\), define the metric\[D(x, y) = \sum_{n=1}^\infty \frac{1}{2^n} \bar{d}_n(x_n, y_n).\]

2.4. The metric \(D\) induces the product topology on \(X\).

2.5. Let \((x^{(k)})_{k=1}^\infty\) be a Cauchy sequence in \((X, D)\).

2.6. For each fixed coordinate \(n\), \(\bar{d}_n(x_n^{(k)}, x_n^{(m)}) \le 2^n D(x^{(k)}, x^{(m)}) \to 0\) as \(k, m \to \infty\), so the coordinate sequence \((x_n^{(k)})_{k=1}^\infty\) is Cauchy in \((X_n, d_n)\).

2.7. By completeness of each \((X_n, d_n)\), \(x_n^{(k)} \to y_n \in X_n\) as \(k \to \infty\).

2.8. Setting \(y = (y_1, y_2, \dots) \in X\), \(x^{(k)} \to y\) in the product topology, so \((X, D)\) is complete.

3.1. Let \((X, d)\) be a complete metric space and let \(U \subseteq X\) be an open subset.

3.2. If \(U = X\), \(U\) is already complete. If \(U \subsetneq X\), the closed set \(F = X \setminus U\) is non-empty.

3.3. Define \(\phi: U \to \mathbb{R}\) by \(\phi(x) = \frac{1}{d(x, F)}\). Since the distance function \(x \mapsto d(x, F)\) is continuous and non-zero on \(U\), \(\phi\) is continuous.

3.4. Define the embedding \(f: U \to X \times \mathbb{R}\) by \(f(x) = (x, \phi(x))\). The map \(f\) is a homeomorphism of \(U\) onto its image \(f(U) \subset X \times \mathbb{R}\) (with inverse being the first projection \(\pi_1\)).

3.5. By parts (a) and (b), the product space \(X \times \mathbb{R}\) is complete under the metric \(D((x_1, t_1), (x_2, t_2)) = d(x_1, x_2) + |t_1 - t_2|\).

3.6. We claim that the graph \(f(U) = \{(x, \phi(x)) : x \in U\}\) is a closed subset of \(X \times \mathbb{R}\):Let \((x_k, \phi(x_k)) \to (x, t)\) in \(X \times \mathbb{R}\) as \(k \to \infty\). Then \(x_k \to x\) in \(X\) and \(\phi(x_k) \to t\) in \(\mathbb{R}\).If \(x \in F\), then \(d(x_k, F) \to d(x, F) = 0\), so \(\phi(x_k) = \frac{1}{d(x_k, F)} \to \infty\), which contradicts \(\phi(x_k) \to t \in \mathbb{R}\).Thus \(x \in U\), and by continuity of \(\phi\), \(t = \lim \phi(x_k) = \phi(x)\).

3.7. Hence \((x, t) = (x, \phi(x)) \in f(U)\), proving that \(f(U)\) is closed in \(X \times \mathbb{R}\).

3.8. By part (a), \(f(U)\) is complete, so \(U \cong f(U)\) is topologically complete.

4.1. Let \(A \subseteq X\) be a \(G_\delta\) subset in a topologically complete space \(X\), so \(A = \bigcap_{n=1}^\infty U_n\) for open sets \(U_n \subseteq X\).

4.2. By part (c), each open set \(U_n\) is topologically complete.

4.3. By part (b), the countable product \(Y = \prod_{n=1}^\infty U_n\) is topologically complete.

4.4. Define the diagonal map \(\Delta: A \to Y\) by \(\Delta(a) = (a, a, a, \dots)\).

4.5. The map \(\Delta\) is a homeomorphism of \(A\) onto its image \(\Delta(A) \subset Y\).

4.6. The image is \(\Delta(A) = \left\{(u_1, u_2, \dots) \in \prod_{n=1}^\infty U_n : u_1 = u_2 = u_3 = \cdots \right\} = \bigcap_{n=1}^\infty \{(u_k) \in Y : u_n = u_{n+1}\}\).

4.7. Since each coordinate equality condition \(\{ (u_k) \in Y : u_n = u_{n+1} \}\) is closed in the product topology (as \(X\) is Hausdorff), \(\Delta(A)\) is an intersection of closed sets, hence closed in \(Y\).

4.8. By part (a), the closed subspace \(\Delta(A) \subseteq Y\) is topologically complete, so \(A \cong \Delta(A)\) is topologically complete.

5.1. The real line \(\mathbb{R}\) with its standard metric is complete, hence topologically complete.

5.2. The rationals \(\mathbb{Q} = \{q_1, q_2, q_3, \dots\}\) are countable, so \(\mathbb{Q} = \bigcup_{n=1}^\infty \{q_n\}\) is an \(F_\sigma\) set.

5.3. The irrational numbers are the complement \(\mathbb{R} \setminus \mathbb{Q} = \bigcap_{n=1}^\infty (\mathbb{R} \setminus \{q_n\})\).

5.4. Since each \(\mathbb{R} \setminus \{q_n\}\) is open in \(\mathbb{R}\), \(\mathbb{R} \setminus \mathbb{Q}\) is a \(G_\delta\) set in \(\mathbb{R}\).

5.5. By 4, the irrational numbers \(\mathbb{R} \setminus \mathbb{Q}\) are topologically complete.