Timeline for On projectively countable sets in the Hilbert cube
Current License: CC BY-SA 4.0
7 events
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| May 1, 2019 at 6:31 | comment | added | Arno | Thanks for pointing out the obvious that I missed - I had taken the scenic route there. I suspect that the computability-theoretic techniques I'd use for this will not work then, but this certainly is an intriguing question. | |
| May 1, 2019 at 5:36 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added Theorem 2
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| May 1, 2019 at 5:21 | comment | added | Taras Banakh | @Arno Each projectively countable subset $A$ of the Hilbert cube $[0,1]^\omega$ is contained in the countable power $C^\omega$ of some countable set $C\subset[0,1]$ and consequently, $A$ is zero-dimensional (not only countably-dimensional). On the other hand, for any compact finite-dimensional subset $K\subset [0,1]^\omega$ the intersection $K\cap A$ is countable. But all this information is insufficient for concluding that $A$ is countable. Moreover, I strongly suspect that under CH the Hilbert cube does contain an uncountable projectively countable subset, put the proof escapes. | |
| May 1, 2019 at 3:34 | comment | added | Arno | I think I know how to show that every projectively-countably subset of the Hilbert Cube is countably-dimensional (so in particular, if there is an uncountable one, then there is an uncountable zero-dimensional projectively countable subset). Would this help? | |
| Apr 30, 2019 at 16:11 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 480 characters in body
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| Apr 30, 2019 at 16:05 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 480 characters in body
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| Apr 29, 2019 at 5:19 | history | asked | Taras Banakh | CC BY-SA 4.0 |