MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a completely metrizable space, say that it has property X if it can be embedded in some metric space such that its image is not closed. For example, the real line R can be embedded, topologically, in itself as (0,1) which is not closed. A compact space such as S^1, however, clearly cannot be embedded in any metric space in this way.

  1. Is it true that a space has property X iff it can be embedded in itself in this way? My intuition says no, but I can't think of a counterexample offhand, partly because I don't know any interesting non-compact metric spaces.

  2. Is property X equivalent to not being compact? If every non-compact (completely metrizable) space has a metrizable compactification then this is easy, but I don't know whether that is the case, although I suspect it might be. Which leads onto the following weaker question:

  3. Given a space which does not have property X, can it be embedded in a space which does? Here my intuition says yes, this should be the case, but I can't think of a more general approach for constructing a suitable embedding space than compactification which, as I said, I can't prove will give a metrizable space. I don't think I know enough topology (I've only taken a couple of basic undergraduate courses).

Edit: 3. above should read "given a space which has property X, can it be embedded in a space which does not?

share|cite|improve this question
A metrizable space has a metrizable compactification if and only if it's second-countable (equivalently, separable). The direction you care about is clear, since any subspace of a compact metric space is second-countable. – Qiaochu Yuan Apr 21 '10 at 5:38
A metrizable topological space S is compact if and only if every metric on S that induces this topology is complete. – Gerald Edgar Apr 21 '10 at 10:32
Thanks to everyone who's contributed, you've all been very helpful. I'm embarrassed to admit it didn't occur to me that 2 might not imply 3 - I'm still not used to thinking of infinite chains of inclusion. Thanks again! – Robin Saunders Apr 21 '10 at 20:25
up vote 6 down vote accepted

The answer to 2 is yes. Let $(X,d)$ be complete but not compact, then for some $r>0$ it has a countable r-separated subset $S=\{s_i\}:i=1,2,\dots$. Let $d'$ be the maximal metric on $X$ satisfying $d'\le d$ and $d'(s_i,s_j)\le r/\min(i,j)$ for all $i,j$. Then $(X,d')$ is homeomorphic to $(X,d)$ - in fact, the identity map is a local isometry - but $d'$ is not complete. So we have embedded the space into the completion of $(X,d')$ as a non-closed subset.

The metric $d'$ can be constructed explicitly: $d'(x,y)$ is the minimum of $d(x,y)$ and the infimum of sums $d(x,s_i)+d(x,s_j)+r/\min(i,j)$ over all pairs of $i,j$. Verifying the triangle inequality is straightforward.

As for 3, the answer is no, because you cannot embed any complete space into a compact space. For example, a non-separable Banach space cannot be so embedded, as Qiaochu Yuan explained in comments.

Update. It seems that I misunderstood Q3. As stated, it asks whether every compact space can be embedded into a complete non-compact one. The answer is of course yes, as Ady noticed.

share|cite|improve this answer
Sorry, the question as you interpreted it was actually the question I'd meant to ask! I've corrected it above. But thanks to Ady for your answer - sadly, any apparent good intuition on my part is purely coincidental. – Robin Saunders Apr 27 '10 at 2:30

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.

It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact and that $d$ is bounded, by taking into consideration the equivalent metric $d^{*}\left(x,y\right)$= $\sup_{t\in M}$ $\mid f(x)\cdot d(x,t)$ $-f\left(y\right)\cdot d\left(y,t\right)\mid$, where $f:M$ $\rightarrow$ $(0,\infty)$ is bounded and continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.

Consequently, $M$ has "the property X" iff it is non-compact.

Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .

[Note that on Wikipedia you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.

share|cite|improve this answer
The "min metric" is not always a metric. – Sergei Ivanov Apr 29 '10 at 8:36
@Sergei Ivanov Of course you have right. Thank you. I've modified my answer. – Ady May 2 '10 at 2:14

The answer to 1. is definitely no: a regular tree of valency at least $3$ is a counter-example. An embedding into itself must have closed image due to the combinatorial constraints. However, it can easily be embedded into a plane with non-closed image.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.