Theorem. Let \(f: X \to Y\); let \(Y\) be compact Hausdorff. Then \(f\) is continuous if and only if the graph of \(f\),
\[ G_f = \ts{x \times f(x) \mid x \in X}, \]
is closed in \(X \times Y\). [Hint: If \(G_f\) is closed and \(V\) is a neighborhood of \(f(x_0)\), then the intersection of \(G_f\) and \(X \times (Y - V)\) is closed. Apply Exercise 7.]
Solution
Goal: Prove the Closed Graph Theorem for topological spaces: if \(Y\) is compact Hausdorff, then \(f: X \to Y\) is continuous if and only if its graph \(G_f \subseteq X \times Y\) is closed.
1. Direct implication (\(\implies\)): If \(f\) is continuous and \(Y\) is Hausdorff, then \(G_f\) is closed in \(X \times Y\).Proof:
1.1. Let \((x_0, y_0) \in (X \times Y) \setminus G_f\), so \(y_0 \neq f(x_0)\).
1.2. Since \(Y\) is Hausdorff, there exist disjoint open sets \(V, W \subseteq Y\) with \(f(x_0) \in V\), \(y_0 \in W\), and \(V \cap W = \emptyset\).
1.3. Since \(f\) is continuous, \(U = f^{-1}(V)\) is an open neighborhood of \(x_0\) in \(X\).
1.4. The set \(U \times W\) is an open neighborhood of \((x_0, y_0)\) in \(X \times Y\).
1.5. For any \((x, y) \in U \times W\), \(x \in U \implies f(x) \in V\), while \(y \in W\). Since \(V \cap W = \emptyset\), \(y \neq f(x)\), so \((x, y) \notin G_f\).
1.6. Thus \((U \times W) \cap G_f = \emptyset\), proving that \((X \times Y) \setminus G_f\) is open. Hence \(G_f\) is closed.
2. Projection along a compact factor is a closed map:If \(Y\) is compact, the projection map \(\pi_1: X \times Y \to X\) defined by \(\pi_1(x, y) = x\) is a closed map.Proof (Tube Lemma):
2.1. Let \(C \subseteq X \times Y\) be closed. We show that \(X \setminus \pi_1(C)\) is open in \(X\).
2.2. Let \(x_0 \in X \setminus \pi_1(C)\), so \((\{x_0\} \times Y) \cap C = \emptyset\), meaning \(\{x_0\} \times Y \subseteq (X \times Y) \setminus C\).
2.3. Since \(Y\) is compact and \((X \times Y) \setminus C\) is open, the Tube Lemma guarantees the existence of an open neighborhood \(U\) of \(x_0\) in \(X\) such that \(U \times Y \subseteq (X \times Y) \setminus C\).
2.4. Then \((U \times Y) \cap C = \emptyset\), so \(U \cap \pi_1(C) = \emptyset\), which means \(U \subseteq X \setminus \pi_1(C)\).
2.5. Thus \(X \setminus \pi_1(C)\) is open, so \(\pi_1(C)\) is closed in \(X\).
3. Reverse implication (\(\impliedby\)): If \(G_f\) is closed in \(X \times Y\) and \(Y\) is compact Hausdorff, then \(f\) is continuous.Proof:
3.1. Let \(x_0 \in X\) and let \(V \subseteq Y\) be an open neighborhood of \(f(x_0)\).
3.2. The complement \(Y \setminus V\) is closed in \(Y\), and since \(Y\) is compact, \(Y \setminus V\) is compact.
3.3. The cylinder \(X \times (Y \setminus V)\) is closed in \(X \times Y\).
3.4. The intersection \(C = G_f \cap (X \times (Y \setminus V))\) is closed in \(X \times Y\).
3.5. By 2, \(\pi_1(C)\) is closed in \(X\).
3.6. We identify the set \(\pi_1(C)\):\[\pi_1(C) = \{x \in X : (x, f(x)) \in X \times (Y \setminus V)\} = \{x \in X : f(x) \in Y \setminus V\} = X \setminus f^{-1}(V).\]
3.7. Since \(\pi_1(C) = X \setminus f^{-1}(V)\) is closed in \(X\), its complement \(f^{-1}(V)\) is open in \(X\).
3.8. Moreover, \(f(x_0) \in V \implies x_0 \in f^{-1}(V)\), so \(f^{-1}(V)\) is an open neighborhood of \(x_0\).
3.9. Thus \(f\) is continuous at \(x_0\). Since \(x_0 \in X\) was arbitrary, \(f\) is continuous on \(X\). Q.E.D.