Every metric space $M$ has a completion $\widehat M$, that is, it can be embedded as a dense subset in a complete metric space. When I first came across this theorem, I thought "well, that's amazing, we don't need equivalence classes of Cauchy sequences or Dedekind cuts anymore, we can use this to construct the real line!" It took me some time to realize that the whole theory of metric spaces is built around the concept of real numbers, so no, you can't use this to build the real line.

Now, can one build up the theory of metric spaces (not all, but just enough to squeeze real numbers out of the rationals) without needing the reals first? "Oh, that's easy, just build the reals from the rationals using Cauchy equivalence classes, that's the basic theory you want." Ok, but can it be generalized? Banach completion is close to what I want, but (i) it's still kind of limited (only applies to vector spaces); and (ii) it works for vector spaces over the field of.... the real numbers! So, again, we need to build the reals first.

Is it just that the real line is our canonical example/definition of "complete" or "continuum", so that everything "complete" or "continuum" needs to be defined on top of it? I would like to know what kind of theorems and/or arguments exist around the idea of completing "incomplete" spaces from scratch.

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    $\begingroup$ I don't know, whether this is relevant (don't know anything about metric spaces), but taking $\mathbb{R}$ to be the codomain / range of a metric is sort of canonical, since it is the maximal archimedian ordered field. (i.e. every other achimedian ordered field is a subfield of $\mathbb{R}$). Non-archimedian ordered fields on the other hand don't really capture our intuition of distance. On a side note: You can also construct the reals from the surreals ;) . $\endgroup$ Commented Feb 28, 2015 at 8:07
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    $\begingroup$ Am I mistaken if I say that $\mathbb Q$ is the source of an infinite number of normed spaces, where the p-adic norms are defined $||x||_p$ for $p=$ primes and even for $p=\infty$ which is the easiest case: $||x||_{\infty}=|x|$. So you have norms over $\mathbb Q$. Why don't you use the Banach compl theorem. Or does it require that the norm be real? I've not recall if the theorem at stake uses that the norm is real valued. I just know that, p.ex. in the case of $\mathbb Q$ the norms are $\mathbb Q$ valued. $\endgroup$
    – zoli
    Commented Feb 28, 2015 at 11:51
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    $\begingroup$ Completing incomplete spaces from scratch is exactly what the theory of Dedekind cuts or Cauchy sequences gets you! $\endgroup$ Commented Aug 30, 2020 at 20:55

1 Answer 1


My too-long-for-a-comment 2¢. It is not possible to build a theory of completion (be it of metric or uniform spaces) without using (uncountable $G_\delta$-sets in) $\mathbb R$ as a model. Results throughout are taken from A. S. Kechris, Classical Descriptive Set Theory, 1995.

  1. Let us restrict our attention to separable metric (rather: metrizable, see below) spaces: if we cannot understand completion of these spaces without $\mathbb R$, then a fortiori we cannot do so for more general spaces (e.g. non-separable uniformizable).

  2. It is not so relevant (to a negative answer to) the question, whether we can realize completions explicitly, or rather whether we can consider the question of the existence of complete metrics metrizing a given topology. In other words, the question in the op is intimately related to

Question: Can we build a theory of Polish spaces without $\mathbb R$?

(Recall that a Polish space is a separable metrizable topological space $(X,\tau)$ so that there exists a distance $d$ metrizing $\tau$ and additionally so that $(X,d)$ is complete as a metric space). The connection to the question in the op is a consequence of 1.: indeed, if $(X,d)$ is separable, then its completion $\overline X$ is complete, thus, in particular, Polish.

  1. The advantage of using Polish spaces comes from the fact that they are at the foundation of descriptive set theory (but I will try to stay away from it as much as possible).

  2. Once we have recast the problem as in 2., then we start seeing why $\mathbb Q$ is not a good model.

Fact 1: $\mathbb Q$ is not a Polish space.

That is, there exists no distance on $\mathbb Q$ (metrizing its standard topology as a subset of the real line and) turning it into a complete metric space. This is a topological fact, but topological facts on 'small' spaces have usually some connection with problems of cardinality. Indeed, we also have

Fact 2: Every countable metrizable space with no isolated points is homeomorphic to $\mathbb Q$.

Combining Facts 1 and 2 we see that a theory of infinite (in cardinality) Polish spaces requires uncountable sets in a very essential way.

  1. Recall that a $G_\delta$-set is any up-to-countable intersection of open sets. $G_\delta$-sets are relevant to the theory of Polish spaces because of the following

Theorem (Alexandrov): A subspace of a Polish space is itself Polish if and only if it is a $G_\delta$-set.

  1. Once $\mathbb Q$ is out of the way, let us turn to why $\mathbb R$ is necessary and sufficient.

Fact 3: $\mathbb R^\mathbb N$, the Hilbert cube $\mathcal H:=[0,1]^\mathbb N$, the Baire space $\mathcal N=\mathbb N^\mathbb N$, and the Cantor space $\{0,1\}^\mathbb N$ are Polish spaces. They are the prototypical Polish spaces in the sense that they satisfy some universal property along the following lines:

  1. (Kechris 4.17) Every Polish space is homeomorphic to a closed subset of $\mathbb R^\mathbb N$;
  2. (Kechris 4.14) Every Polish space is homeomorphic to a $G_\delta$-set in $\mathcal H$;
  3. I am not listing the properties of the Baire space and of the Cantor space since they are not relevant in the discussion below.

Fact 3.1. is the relevant one here: since every closed subset of a complete space is complete, every Polish space is homeomorphic to a complete metric subspace of $\mathbb R^\mathbb N$, i.e. one that is complete with a particular assigned metric. A proof of both Facts 3.1 and 3.2 relies on the completeness of $\mathcal H$ and, implicitly, on that of the unit interval, thus again on the completeness of $\mathbb R$.

  1. Let us turn back to the problem of completion. Metric (and uniform) completions are characterized by the following universal property:

Fact 4 (completion): For every metric space $(X,d)$ there exists a metric space $(\overline X,\overline d)$ (unique up to isomorphism) and a dense embedding $\iota\colon X\to\overline X$ with the following property. For every complete metric space $(Y,\rho)$ and every uniformly continuous map $f\colon X\to Y$ there exists a unique (uniformly) continuous extension $\overline f\colon \overline X\to Y$ so that $f=\overline f \circ \iota$.

While seemingly unrelated to Polishness or $G_\delta$ sets, Fact 4 may be understood as the analogon in $\mathbf{Met}$ of the following result in $\mathbf{Top}$:

Theorem (Kuratowski, Kechris 3.8) Let X be metrizable and Y completely metrizable. Further let $A\subset X$ and $f\colon A\to Y$ be continuous. Then there is a $G_\delta$ set $G$ with $A\subset G\subset \overline A$ and a continuous extension $f_G$ of $f$ to $G$.

In turn, Kuratowski's Theorem is related to the realization of Polish spaces as subspaces of $\mathcal H$ by the following result of Lavrentiev.

Theorem (Lavrentiev, Kechris 3.9) Let $X,Y$ be completely metrizable, $A\subset X$, $B\subset Y$, and $f\colon A\to B$ be a homeomorphims. Then there exist $G_\delta$-sets $G\supset A$ and $H\supset B$ and a homeomorphism $h\colon G\to H$ extending $f$.

Note. While the above line of reasoning is slightly more implicit than necessary, this is essentially due to the additional of requirement of separability, which makes things more understandable but somewhat less polished.

If we drop separability and metrizability and we only understand completeness in the correct category $\mathbf{USp}$ of uniform spaces with uniform maps, then we have the following result of Shirota, which streamlines the above discussion:

Theorem (Shirota, Osaka J. Math., 1952) For a completely regular Hausdorff topological space, the following are equivalent:

  • the natural uniformity of $X$ is complete;
  • X is realcompact (equivalently: homeomorphic to a closed subset of a Cartesian product of real lines).

(See also A.6.10 in Borchers–Sen, Mathematical Implications of Einstein-Weyl Causality, 2006)

Again, completeness is characterized by the property of being a closed subspace of a product of real lines, as in Fact 3.1.

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    $\begingroup$ Only now I had the time to read this thoroughly. That's awesome, thank you for posting! I'm marking this as accepted because I don't believe I'm getting an answer any better than that. $\endgroup$
    – fonini
    Commented Sep 10, 2022 at 18:51
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    $\begingroup$ Thank you. For me this is more or less the reason (the topological one), but I can imagine how a more abstract (e.g. category theory/descriptive set theory) approach might provide further insights in possibly different directions. $\endgroup$
    – AlephBeth
    Commented Sep 10, 2022 at 22:21
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    $\begingroup$ Wow, some cool facts I had not known! Thanks! :) $\endgroup$ Commented Jan 17, 2023 at 2:41

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