4
$\begingroup$

This is homework and I'm not looking for an answer. I just finished an exercise that asked to prove that every metric compact space can be embedded in the Hilbert cube. Knowing this I can see that I have to find a non-compact metric space to start, but I don't think that's enough because I've been thinking for a while now and nothing has come to mind, so a little intuition or hints on what I'm looking for would be appreciated. Thanks.

$\endgroup$
  • $\begingroup$ Have you already learned about first- and second-countable spaces? $\endgroup$ – Daniel Fischer Sep 12 '15 at 20:16
  • $\begingroup$ I have, but I don't see how to apply it. $\endgroup$ – user178318 Sep 12 '15 at 20:42
3
$\begingroup$

HINT: It isn’t really compactness that matters here: it’s second countability. You want a metric space that isn’t second countable; equivalently, you want one that isn’t separable. There are some really simple non-separable metric spaces. A further hint is in the spoiler-protected block below.

Try discrete spaces.

$\endgroup$
  • $\begingroup$ I though about using the real numbers with the discrete topology. I see clearly that this space is not second countable since the singletons are open in this topology, and it's metric using d∞. I don't think though it's completely clear to me why second countability is what we need. Could it be possible to prove that a metric second-countable space can be embedded in the Hilbert cube? Also, I proved some exercises ago that every metric compact space is separable and even second countable. Am I wrong to think the space has to be T4 so it can be embedded in the Hilbert cube? $\endgroup$ – user178318 Sep 12 '15 at 21:10
  • $\begingroup$ @416256: That example will work. I don’t know how you proved that every compact metric space embeds in the Hilbert cube, but one common argument actually shows that every second countable $T_3$-space does: one uses regularity and second countability to get a countable family of continuous functions to $[0,1]$ that distinguish points and then uses them to define the embedding. This proof makes it clear how second countability is what’s needed. \\ The Hilbert cube is metrizable, so every subspace is metrizable and hence normal; thus, anything that embeds in the Hilbert cube must indeed be $T_4$. $\endgroup$ – Brian M. Scott Sep 12 '15 at 21:17
  • $\begingroup$ I think I get it now. You could then say it's equivalent to ask for a space X that can be embedded in the Hilbert cube, and to ask for a second countable regular space? I´ll try to prove both implications. I was confused because I first thought it was only one way and didn't know hot to prove my example actually worked, but if I manage to prove the equivalence then I'll be done. Thanks a lot, you clarified many things. $\endgroup$ – user178318 Sep 12 '15 at 21:45
  • $\begingroup$ @416256: You’re very welcome. Yes, that’s right. You know that regularity is hereditary, so you just have to show that every subspace of a second countable space is second countable, which is pretty easy to do. $\endgroup$ – Brian M. Scott Sep 12 '15 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy