Questions: Metrization Theorems

Question 1

A mathematician defines a topology on a set by specifying which subsets are open. She wants to use Cauchy sequences and completeness in her analysis. Can she proceed immediately?

• A. Yes — every topological space has an underlying metric that defines convergence
• B. Yes — Cauchy sequences are defined using only convergence, which is available in any topological space
• C. No — she must first verify the space is metrizable by checking conditions like second-countability and regularity
• D. No — completeness is never available in topological spaces, only in explicit metric spaces

Answer: (C)

Not every topological space is metrizable. Cauchy sequences and completeness require a metric for their definition — 'the distance between terms eventually becomes less than ε' has no meaning without a notion of distance. She must first verify the space satisfies the hypotheses of a metrization theorem (e.g., second-countable + regular Hausdorff for Urysohn's theorem) to confirm a metric exists. Option D is too strong — metrizable spaces are complete if and only if that metric is complete.

Question 2

What does the Urysohn Metrization Theorem assert?

• A. Every Hausdorff space can be given a metric compatible with its topology
• B. A topological space is metrizable if and only if it has a σ-locally finite basis
• C. Every second-countable, regular Hausdorff space is metrizable
• D. Every metric space is second-countable and regular

Answer: (C)

Urysohn's theorem: second-countable + regular + Hausdorff implies metrizable. Option A is false — many Hausdorff spaces fail to be metrizable (e.g., the long line). Option B is the Nagata–Smirnov theorem, a more general result that characterizes all metrizable spaces. Option D describes what metrizable spaces must satisfy but is not the theorem's direction.

Question 3

A metrizable topological space must be Hausdorff, since any two distinct points can be separated by open balls.

• T. True
• F. False

Answer: True

True — in any metric space, given distinct points x and y with d(x,y) = r > 0, the open balls B(x, r/2) and B(y, r/2) are disjoint open sets separating them. Hausdorff separation is therefore a necessary condition for metrizability, and one reason why non-Hausdorff topologies (like the Zariski topology on algebraic varieties) are not metrizable.

Question 4

A metrization theorem guarantees a unique metric for a given topological space — that is, at most one metric generates any given topology.

• T. True
• F. False

Answer: False

False — metrization is an existence result, not a uniqueness result. Many different metrics can generate the same topology. For example, on ℝⁿ, the Euclidean metric, the taxicab metric, and the max metric all generate the same (standard) topology. The Urysohn proof constructs one particular metric, but it is not the only one possible.

Question 5

Why does second-countability play a crucial role in proving that a regular Hausdorff space is metrizable?

Model answer: Second-countability provides 'size control' that allows the construction of a countable family of continuous functions separating points. The proof uses the Urysohn lemma (requiring normality, which follows from regularity plus second-countability) to build these functions, then defines a metric as a weighted combination of them. Without second-countability, this countable construction fails — the Nagata–Smirnov theorem replaces second-countability with the weaker σ-locally finite basis condition for the fully general result.

The key insight is that metrization requires 'enough' continuous real-valued functions to distinguish all points — the real line's geometry must be embeddable into the abstract space. Second-countability, by controlling the size of the topology, ensures such a countable separating family exists. This is why, intuitively, metrizable spaces must not be 'too large' in topological terms.