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If $X$ is any metrizable space, $A$ is a closed subset of $X$. Let $d$ be a compatible metric on $A$ then $d$ can be extended to a compatible metric on $X$.

Hagen von Eitzen
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akansha
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2 Answers2

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I cite (with a correction) the beginning of my paper “On Extension of (Pseudo-)Metrics from Subgroup of Topological Group onto the Group”:

“The problem of extensions of functions from subobjects to objects in various categories was considered by many authors. The classic Tietze-Urysohn theorem on extensions of functions from a closed subspace of a topological space and its generalizations belong to the known results. Hausdorff [3] showed that every metric from a closed subspace of a metrizable space can be extended onto the space. Isbell [4, Lemma 1.4] showed that every bounded uniformly continuous pseudometric on a subspace of a uniform space can be extended to a bounded uniformly continuous pseudometric on the whole space. The linear operators extending metrics from a closed subspace of a metrizable space onto the space were considered in, e.g., [2,6]”.

References

[2] Bessaga C., Functional analytic aspects of geometry. Linear extending of metrics and related problems, in: Progress of Functional Analysis, Proc. Peniscola Meeting 1990 on the 60th birthday of Professor M. Valdivia, North-Holland, Amsterdam (1992) 247-257.

[3] Hausdorff F. Erweiterung einer Homömorpie, - Fund. Math., 16 (1930) 353-360.

[4] Isbell J.R. On finite-dimensional uniform spaces, - Pacific J. of Math., 9 (1959) 107-121.

[6] Zarichnyi M., Regular Linear Operators Extending Metrics: a Short Proof, Bull. Pol. Ac.:Math., 44 (1996) 267-269.

Alex Ravsky
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    Re :"Isbell [4] showed that..." This doesn't look right.What if the subspace is empty? Not all uniform space are metrizable. – DanielWainfleet Sep 10 '15 at 20:44
  • @DanielWainfleet Thanks. I corrected it and added a link to the paper. – Alex Ravsky Jan 04 '19 at 14:07
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    @AlexRavsky Hello! I was wondering the following. Let X be a uniform space and A dense subspace (with the uniform topology). Suppose that A has a (bounded) metric $d$. Is it possible to extend it to a metric on X? .....Not a pseudo-metric but a metric. Thanks! – Curious Jul 20 '20 at 11:39
  • @AlexRavsky To be precise, $d$ is uniformly continuous on A. – Curious Jul 20 '20 at 12:28
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    @Curious If the extension should be uniformly continuous then the answer can be negative. Indeed, let $X$ be the unit segment $[0,1]$ endowed with the usual uniformity, $A=(0,1)$ and $d(x,y)=|e^{2\pi x}- e^{2\pi y}|$. Then $d$ is uniformly continuous on $A\times A$, but $\bar d(0,1)=0$ for any continuous extension $\bar d$ of $d$ to $X^2$. – Alex Ravsky Aug 08 '20 at 04:57
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    @AlexRavsky, thanks :) – Curious Aug 08 '20 at 17:38
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A short accessible proof of Hausdorff's metric extension theorem can be also found in:

Torunczyk, H., A short proof of Hausdorff’s theorem on extending metrics, Fundam. Math. 77, 191-193 (1973). ZBL0248.54035.

The author includes several references with alternative proofs of the theorem.

Moishe Kohan
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