Timeline for Is there a compact group of countably infinite cardinality?
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| Dec 9, 2009 at 9:54 | comment | added | Thorny | Right, this may not end up with a set homeomorphic to the Cantor set, but rather one with a surjection onto a Cantor set. Let $A^{s0}_n$ and $A^{s1}_n$ ($s$ is a length $n-1$ bit string) be the two disjoint closed sets within $A^{s}_{n-1}$, with the starting set $A^{\emptyset}_0$, the whole space. For each chain $A^{b_1}_1 \supset A^{b_1b_2}_2 \supset ...$ of closed (hence compact) sets their intersection will be nonempty; in general, it may not be a single point, but mapping it to the corresponding point $\sum \frac{b_j}{3^j}$ in the Cantor set will give you a surjective (and continuous) map. | |
| Dec 8, 2009 at 17:30 | comment | added | Yemon Choi | I like the idea, but is it really that easy to show that the construction gets you a big enough set? In a metric space I agree that this intersection will contain a copy of the Cantor set, but in a general LCH space I seem to be missing something. (I'm not saying the construction doesn't work, but that it seems to require a bit more slogging than your answer suggests.) | |
| Dec 8, 2009 at 15:02 | history | answered | Thorny | CC BY-SA 2.5 |