Let \(X\) be metrizable. Show that the following are equivalent:
\(X\) is bounded under every metric that gives the topology of \(X\).
Every continuous function \(\phi: X \to \mathbb{R}\) is bounded.
\(X\) is limit point compact.
[Hint: If \(\phi: X \to \mathbb{R}\) is a continuous function, then \(F(x) = x \times \phi(x)\) is an imbedding of \(X\) in \(X \times \mathbb{R}\). If \(A\) is an infinite subset of \(X\) having no limit point, let \(\phi\) be a surjection of \(A\) onto \(\mathbb{Z}_+\).]
Solution
Goal: Prove the equivalence of (i) universal metric boundedness, (ii) pseudocompactness (boundedness of continuous real-valued functions), and (iii) limit point compactness for any metrizable space \(X\).
1. (i) \(\implies\) (ii):Proof:
1.1. Suppose for contradiction that there exists a continuous function \(\phi: X \to \mathbb{R}\) that is unbounded.
1.2. Let \(d_0\) be an arbitrary compatible metric on \(X\).
1.3. Define a new function \(d: X \times X \to [0, \infty)\) by:\[d(x, y) = d_0(x, y) + |\phi(x) - \phi(y)|.\]
1.4. \(d\) is a metric on \(X\): positivity, non-degeneracy, symmetry, and the triangle inequality hold immediately from the properties of \(d_0\) and the absolute value metric on \(\mathbb{R}\).
1.5. \(d\) generates the original topology of \(X\): the map \(F(x) = (x, \phi(x))\) is a topological embedding of \(X\) into \((X, d_0) \times \mathbb{R}\) with the \(\ell_1\) product metric \(d((x_1, t_1), (x_2, t_2)) = d_0(x_1, x_2) + |t_1 - t_2|\).
1.6. Since \(\phi(X)\) is unbounded in \(\mathbb{R}\), \(\operatorname{diam}_d(X) \ge \sup_{x, y \in X} |\phi(x) - \phi(y)| = \infty\).
1.7. Thus \(X\) is unbounded under the compatible metric \(d\), contradicting (i).
1.8. Therefore every continuous function \(\phi: X \to \mathbb{R}\) must be bounded.
2. (ii) \(\implies\) (iii):Proof:
2.1. Suppose for contradiction that \(X\) is not limit point compact.
2.2. There exists an infinite subset \(A \subseteq X\) with no limit points in \(X\).
2.3. Let \(A_0 = \{a_1, a_2, \dots\}\) be a countably infinite subset of \(A\). Since \(A\) has no limit points, \(A_0\) has no limit points, so \(A_0\) is a closed and discrete subspace of \(X\).
2.4. Define \(f: A_0 \to \mathbb{R}\) by \(f(a_n) = n\). Because \(A_0\) is discrete, \(f\) is continuous.
2.5. Since \(X\) is metrizable, \(X\) is normal (\(T_4\)).
2.6. By the Tietze Extension Theorem, \(f\) extends to a continuous function \(\phi: X \to \mathbb{R}\).
2.7. Since \(\phi(a_n) = n\) for all \(n \in \mathbb{Z}_+\), \(\phi\) is unbounded on \(X\), contradicting (ii).
2.8. Thus \(X\) is limit point compact.
3. (iii) \(\implies\) (i):Proof:
3.1. For any metrizable space, limit point compactness is equivalent to compactness.
3.2. Let \(d\) be any metric inducing the topology of \(X\). Since \(X\) is compact, \((X, d)\) is a compact metric space.
3.3. Every compact metric space is bounded: fixing \(x_0 \in X\), the open cover \(\{B_d(x_0, n)\}_{n=1}^\infty\) has a finite subcover, so \(X \subseteq B_d(x_0, N)\), giving \(\operatorname{diam}_d(X) \le 2N < \infty\).
3.4. Thus \(X\) is bounded under \(d\), establishing (i).
4. Conclusion:The conditions (i), (ii), and (iii) are mutually equivalent. Q.E.D.