Let \(X\) be a locally compact space. If \(f: X \to Y\) is continuous, does it follow that \(f(X)\) is locally compact? What if \(f\) is both continuous and open? Justify your answer.
Solution
Goal: Determine whether the continuous image (and continuous open image) of a locally compact space is locally compact.
1. Question 1: Continuous images of locally compact spaces need NOT be locally compact.Proof:
1.1. Let \(X = \mathbb{R}\) with the discrete topology.
1.2. Since every singleton in \(X\) is an open and compact neighborhood of itself, \(X\) is locally compact.
1.3. Let \(Y = \mathbb{Q}\) with the standard subspace topology inherited from \(\mathbb{R}\).
1.4. Let \(f: X \to Y\) be any surjective function (for example, mapping \(\mathbb{R} \setminus \mathbb{Q}\) onto \(0\) and acting as the identity on \(\mathbb{Q}\)).
1.5. Since \(X\) has the discrete topology, \(f\) is continuous, and \(f(X) = \mathbb{Q}\).
1.6. The rational numbers \(\mathbb{Q}\) are not locally compact at any point: every compact subset of \(\mathbb{Q}\) has empty interior, so no point of \(\mathbb{Q}\) possesses a compact neighborhood.
1.7. Thus the continuous image \(f(X)\) is not locally compact.
2. Question 2: If \(f\) is continuous and open, then \(f(X)\) IS locally compact.Proof:
2.1. Let \(y \in f(X)\). Since \(f: X \to f(X)\) is surjective, choose \(x \in X\) such that \(f(x) = y\).
2.2. Because \(X\) is locally compact, there exists a compact subspace \(C \subseteq X\) containing an open neighborhood \(U \subseteq X\) of \(x\), so \(x \in U \subseteq C\).
2.3. Since \(f\) is continuous, the image \(f(C)\) is a compact subset of \(f(X)\).
2.4. Since \(f: X \to f(X)\) is an open map, \(f(U)\) is an open subset of \(f(X)\).
2.5. We have \(y = f(x) \in f(U) \subseteq f(C)\), which exhibits \(f(C)\) as a compact neighborhood of \(y\) in \(f(X)\).
2.6. Since \(y \in f(X)\) was arbitrary, \(f(X)\) is locally compact. Q.E.D.