Metrization and Completeness

A topological space is a set together with a chosen family of "open sets" — nothing more. A metric space is far richer: it comes with an actual ruler, a number d(x, y) for every pair of points. Every metric quietly hands you a topology (declare a set open when it contains a small ball around each of its points), so every metric space is a topological space. The reverse question is the deep one, and it is the whole of this page:

Given a bare topological space, is there any metric at all whose open balls reproduce exactly its open sets? If yes, the space is called metrizable — it was "secretly metric all along", and the entire machinery of Cauchy sequences, completeness and uniform structure becomes available. If no, the space genuinely lives beyond the reach of any ruler.

This is not a curiosity. Metrizability is the bridge between the two great generalisations of calculus — point-set topology on one side, metric-space analysis on the other. Knowing when the bridge exists tells you exactly when you may import \varepsilon–\delta arguments, sequences and completeness into a topological setting. And when we cross that bridge we will meet a startling asymmetry: some of the properties we cherish — like completeness — turn out not to be topological at all. They belong to the ruler, not to the space.

We lean on compactness and the separation axioms throughout, so keep them close at hand.

What a metric forces: the necessary conditions

Before hunting for a metric, notice everything a metric would demand. If a space is metrizable, it cannot escape these consequences — so any space failing one of them is instantly disqualified.

These are genuine hurdles: fail any one and metrization is hopeless. But — and this is the subtle part — clearing all of them is not enough. There are Hausdorff, even normal spaces that are still not metrizable. We need a theorem that tells us when the necessary conditions become sufficient.

Urysohn's metrization theorem

The first clean answer, and one of the jewels of point-set topology, packages "enough separation" together with "not too big" and concludes metrizability. Recall that regular (T3) means a point and a disjoint closed set can be separated by disjoint open sets, and second-countable means the whole topology has a countable base of open sets.

The conditions are lightweight and easy to check, which is why this is the workhorse. Any surface, any manifold, any subspace of \mathbb{R}^n sails through: they are regular Hausdorff and have a countable base (balls of rational radius about rational-coordinate points), so Urysohn hands them a metric for free.

But second-countability is a real restriction — plenty of perfectly reasonable spaces have no countable base. To reach all of them we need the definitive result, which drops the countability cap entirely.

Spaces with no ruler at all

Non-metrizable spaces are not exotic monsters — the simplest ones fail at the very first hurdle, Hausdorffness.

Worked example 1 — the cofinite topology

Let X be an infinite set (say \mathbb{N}) with the cofinite topology: a non-empty set is open exactly when its complement is finite. This really is a topology. Now take two distinct points x \ne y and any open sets U \ni x, V \ni y. Each of U, V omits only finitely many points, so U \cap V omits only finitely many — and since X is infinite, that intersection is never empty. No two non-empty open sets are disjoint! The space is not Hausdorff, so by the necessary conditions above it cannot be metrizable. (In fact any sequence of distinct points converges to every point at once — a scandal no metric would ever permit.)

Worked example 2 — an uncountable product of lines

Metrizability can also fail higher up, on a space that is Hausdorff and normal-ish. Take the product

\mathbb{R}^{\mathbb{R}} = \prod_{t \in \mathbb{R}} \mathbb{R}

— all functions \mathbb{R} \to \mathbb{R}, with the product topology (pointwise convergence). This is Hausdorff, yet it is not first-countable at any point: no countable family of basic neighbourhoods can pin down a point, because each basic open set constrains only finitely many coordinates and there are uncountably many coordinates to control. Failing first-countability, it fails a necessary condition — no metric exists. A countable product of metric spaces, by contrast, is always metrizable (there is a tidy weighted-sum metric), so it is precisely the uncountability that breaks the ruler.

The metric is not unique — and that changes everything

Suppose a space is metrizable. Which metric does it get? Not one — many. Two metrics d and d' on the same set are topologically equivalent if they generate the identical family of open sets; they then agree on every purely topological verdict — which sets are open, which sequences converge, which maps are continuous, whether the space is compact or connected.

For instance d and the truncated metric d'(x, y) = \min\big(1,\, d(x, y)\big) are always topologically equivalent: for small radii the balls are literally the same, so they define the same topology. Yet d' makes the whole space have diameter \le 1 while d may be unbounded. That is the crux:

The classic blunder is to treat completeness like compactness — as if a homeomorphism should carry it along. It does not. Consider the map

f(x) = \tfrac{1}{2}\big(\tanh x + 1\big), \qquad f : \mathbb{R} \to (0, 1),

a continuous bijection with continuous inverse. So \mathbb{R} and the open interval (0, 1) are homeomorphic — topologically the same space. But \mathbb{R} is complete and (0, 1) is not: the sequence 1/2, 1/3, 1/4, \dots is Cauchy in (0, 1) yet its would-be limit 0 has been exiled from the space. A homeomorphism happily maps a convergent sequence to a non-convergent one, because it need not respect distances — only open sets.

The other half of the same trap: not every Hausdorff space is metrizable. Being Hausdorff is necessary, never sufficient — the Sorgenfrey line, the uncountable product above, and the long line are all Hausdorff and all resist every metric. "Hausdorff" buys you the necessary conditions, not the metric.

Completeness, revisited — and completion

Recall from Cauchy sequences that (x_n) is Cauchy when its terms eventually crowd together, d(x_m, x_n) \to 0, and that a metric space is complete when every Cauchy sequence actually converges to a point of the space:

(X, d) \text{ complete} \iff \big[(x_n) \text{ Cauchy} \Rightarrow x_n \to x \in X\big].

The rationals \mathbb{Q} are the standard cautionary tale: the truncations 1,\ 1.4,\ 1.41,\ 1.414,\dots form a Cauchy sequence whose limit \sqrt{2} is missing. \mathbb{Q} is riddled with such holes. The cure is to fill them all in at once.

Apply this to \mathbb{Q} and you get exactly the real line: \mathbb{R} is the completion of \mathbb{Q}. The irrationals are the equivalence classes of rational Cauchy sequences with no rational limit — a construction (due to Cantor) that builds the reals from the rationals with nothing but the metric.

Worked example 3 — same topology, one complete metric and one not

Here is completeness's non-topological nature caught in the act on a single space. On \mathbb{R} compare the usual metric with

\rho(x, y) = \big|\arctan x - \arctan y\big|.

Because \arctan is a homeomorphism of \mathbb{R} onto (-\tfrac{\pi}{2}, \tfrac{\pi}{2}), \rho induces the same open sets as |x - y| — the two metrics are topologically equivalent, so as topological spaces they are indistinguishable. But watch the sequence x_n = n: since \arctan n \to \tfrac{\pi}{2}, the values \arctan n crowd together, so \rho(x_m, x_n) \to 0 — the sequence is Cauchy under \rho. Yet x_n = n runs off to infinity and converges to nothing in \mathbb{R}. So (\mathbb{R}, \rho) is not complete, while (\mathbb{R}, |\cdot|) is. Same topology, opposite verdict on completeness — completeness lives in the ruler.

Surprisingly, no — it depends which ruler you use to measure "small". Fix a prime p and declare a rational small when it is divisible by a high power of p: the p-adic absolute value makes p^{100} tiny and 1/p^{100} huge. This is a genuine metric on \mathbb{Q}, inducing a totally different topology from the usual one. Complete \mathbb{Q} in that metric and you get not \mathbb{R} but the p-adic numbers \mathbb{Q}_p — a different complete field for every prime. Ostrowski's theorem says these (plus the ordinary one giving \mathbb{R}) are essentially the only ways to complete \mathbb{Q}. The gaps you find depend entirely on the ruler you carry, and number theorists have built a whole universe inside \mathbb{Q}_p.

The one-page summary