Completeness belongs to the metric; complete metrizability is characterised on a later page

This page has proved that completeness is a property of the metric and not of the topology it induces: two metrics on one set can have exactly the same open sets while only one of them is complete (FALSE: completeness of a metric space is determined by its topology, Complete metric space: every Cauchy sequence converges in the space, Topologically, uniformly and Lipschitz equivalent metrics on a set). That leaves an obvious question, and this remark says what the question is, what the page now answers, and what it does not.

The question. Given the open sets, is there some metric inducing them that is complete? A topology for which the answer is yes is called completely metrizable, and that is the definition made precise in Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete. Unlike completeness, this really is a property of the open sets alone: the metric is quantified over (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), so a homeomorphism transports it — which is claim 1 of that lemma. It is the topological shadow that completeness casts, and it is strictly weaker than "carries this particular complete metric".

What this page now settles. Two of the three facts below are discharged by Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete; the third is not, and says so.

• $(0,\mathrm{\infty })$ with its usual metric is not complete: the sequence $1\mathrm{/}(k+2)$ is Cauchy there and converges in $\mathbb{R}$ only to $0$, which is outside the space. Claim 3 of Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete also states this conclusion. And yet another metric on $(0,\mathrm{\infty })$, inducing exactly the same open sets, is complete: the same claim writes it down as $\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }1\mathrm{/}x-1\mathrm{/}y\mathrm{\mid }$. So the two notions genuinely differ, and the question above is not a distinction without a difference.
• Under Countable Choice, Closed subspaces of complete metric spaces are complete; the converse under countable choice says a subspace of a complete space is complete precisely when it is closed. Its closed-implies-complete direction holds without choice. Claim 2 of Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete upgrades that to the topological statement: a closed subspace of a completely metrizable space is completely metrizable, with no completeness hypothesis on the ambient metric. What happens for subspaces that are not closed is left open here.
• Under Countable Choice, Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences embeds every metric space densely in a complete one. The completion is a complete space, but the original space usually sits inside it as a proper dense subspace, and being a dense subspace of a complete space says nothing on its own about complete metrizability. $\mathbb{Q}$ and $(0,\mathrm{\infty })$ are both dense in complete spaces and they differ on the property: $(0,\mathrm{\infty })$ has it by the first bullet, and $\mathbb{Q}$ does not — but that second half is not proved here and needs the Baire category theorem.

What this early page does not prove. The later published page complete-metrizability-and-baire proves the qualified Alexandrov characterisation: assuming Dependent Choice, a subspace of a complete metric space is completely metrizable exactly when it is a $G_{\delta }$ subset (Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is $G_{\delta }$ ↗). Its published subspace corollary proves the open and closed cases directly, and the general $G_{\delta }$ case under DC (Open and closed subspaces of a completely metrizable space are completely metrizable, and under Dependent Choice so is every $G_{\delta }$ subspace ↗). This early page has not yet developed countable intersections of open sets, the Baire category theorem, or the convergent-series remetrisation used later; The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement here supplies only the metric topology as a collection of subsets. Beyond the three claims of Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete, "completely metrizable" is not used here as a proof tool.

How to read the rest of the library in the meantime. Every statement of the form "$X$ is complete" in this library is a statement about a named metric on $X$, and it never means "$X$ has a complete metric". Where the distinction matters, the metric is written out. This is the same discipline as for the word bounded, which is also metric and not topological (Topologically, uniformly and Lipschitz equivalent metrics on a set).

• This item proves nothing and is not cited by any proof. It records what the page has and has not established, and points at where the missing part will be developed. It is included because the gap it names is the single most common place where a reader over-reads FALSE: completeness of a metric space is determined by its topology: from "completeness is not topological" it does not follow that no topological invariant is in the neighbourhood.
• Forward-reference bookkeeping. The part proved on this page is an ordinary same-page dependency on Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete. The Alexandrov theorem and the open/closed/$G_{\delta }$ subspace corollary are published on the later page and are listed as forward references above. Neither later result is a load-bearing premise for this orientation item.