Find a metric space that cannot be embedded in the Hilbert cube.

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.

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

• @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 $T_4$. – Brian M. Scott Sep 12 '15 at 21:17