Timeline for Infinite topological spaces such that every subset is a retract
Current License: CC BY-SA 3.0
3 events
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| Dec 9, 2014 at 17:53 | comment | added | François G. Dorais | @Eric: Ah yes, good catch! One needs to see that each chain in the specialization order is open. Which is basically what my earlier $T_1$ argument shows as you point out in your answer. | |
| Dec 9, 2014 at 17:44 | comment | added | Eric Wofsey | This argument only proves that the specialization order on $X$ is a subset of $\coprod_i \mathbb{Z}$; you must additionally prove that $X$ has the Alexandrov topology (which I do in my answer). | |
| Dec 9, 2014 at 17:38 | history | answered | François G. Dorais | CC BY-SA 3.0 |