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Nov 13 at 20:12 history edited Jakobian CC BY-SA 4.0
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Nov 11 at 2:16 history edited Jakobian CC BY-SA 4.0
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Nov 11 at 2:10 history edited Jakobian CC BY-SA 4.0
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Nov 11 at 1:49 comment added Anonymous In fact, I wouldn't be surprised if given any countable subset $A$ of $\beta \omega \setminus \omega$, every continuous image of $\beta \omega \setminus A$ is zero-dimensional.
Nov 11 at 1:30 history answered Jakobian CC BY-SA 4.0