Timeline for Zero dimensional iff every closed set is a retract.
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2011 at 23:48 | comment | added | Not Mike | @Brandsma: wasn't meant to a perfect answer, just a fact that came off the top of my head and seemed helpful. That being said, you are correct, I should have added the assumption of compactness, and not having isolated points. | |
| Jan 21, 2011 at 14:28 | answer | added | KP Hart | timeline score: 8 | |
| Jan 21, 2011 at 8:06 | comment | added | Henno Brandsma | @Michael: no, only if there are no isolated points and it is compact. Otherwise Q, R\Q, a convergent sequence etc. are counterexamples. | |
| Jan 20, 2011 at 9:15 | comment | added | Bill Johnson | Problem 1. If $Y$ is a closed subset of a compact metric space $X$ and the nearest point mapping from $X$ to $Y$ is single valued, then it is continuous. Problem 2. Put an equivalent metric on the Cantor set $X=\{0,1\}^N$ so that the hypothesis of (1) is satisfied for every closed subset $Y$ of $X$. Problem 3. Show that Michael and I gave essentially the same hint. | |
| Jan 20, 2011 at 7:40 | comment | added | Not Mike | Hint: Effectively, every zero-dimensional seperable metrizable space is homeomorphic to the cantor middle thirds set on [0,1]. This should be a good starting point for intuition to take over. | |
| Jan 20, 2011 at 7:26 | history | asked | Raj | CC BY-SA 2.5 |