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Timeline for A rare property of Hausdorff spaces

Current License: CC BY-SA 3.0

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Aug 9, 2016 at 21:14 comment added Forever Mozart not so rare, huh?
Aug 9, 2016 at 17:38 comment added Gerald Edgar Locally constant in $\{0,1\}^A$? No. Think of the projection onto $\{0,1\}^B$, where $B \subset A$ is countable. (In fact, the general continuous real-valued function factors through such a projection.) But we do have each nonempty level set $f^{-1}(x)$ has the same power as the whole space.
Aug 9, 2016 at 17:18 comment added Joel David Hamkins Nice example, Gerald. It is similar to my example, but do you get the locally-constant property? It seems not, since the projection functions to the unit interval are continuous, but not locally constant. But if you used $2^A$ instead of $[0,1]^A$, then at least the projection functions would be locally constant. Is every continuous function locally constant in this case?
Aug 9, 2016 at 13:42 history answered Gerald Edgar CC BY-SA 3.0