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Feb 27, 2014 at 23:39 history edited user9072
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Oct 14, 2012 at 17:56 answer added Henno Brandsma timeline score: 4
Aug 22, 2012 at 23:34 comment added Andreas Blass The Appert space, pointed out in Ramiro de la Vega's answer, also fits the description in my previous comment.
Aug 22, 2012 at 19:12 answer added Ramiro de la Vega timeline score: 7
Aug 21, 2012 at 15:34 comment added Andreas Blass In the hope of avoiding an explosion of variants of Stefan's proof, I note that his and Arthur's proofs are both special cases of the following (and there are lots more special cases). Take a countable set $S$ and a filter $F$ on S having no countable base and containing all cofinite subsets of $S$. Topologize $S\cup\{p\}$ (where $p\notin S$) by making a subset $U$ open if either $p\notin U$ or $(U\cap S)\in F$. Then $X$ is normal because of any two disjoint (closed) sets one is open. There is no countable base for the topology because there is no countable neighborhood base at $p$.
Aug 21, 2012 at 5:17 answer added user642796 timeline score: 5
Aug 20, 2012 at 22:20 answer added Stefan Geschke timeline score: 5
Aug 20, 2012 at 21:52 history edited Ramsey CC BY-SA 3.0
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Aug 20, 2012 at 21:09 history asked Pedro Perez CC BY-SA 3.0