A $T1$ space is a topological space where, for any 2 points, each has a neighborhood not containing the other. I wonder if every singleton being closed is an equivalent condition, and what the proof for that is. I also wonder if the following condition is equivalent: for any 2 points $x$ and $y$, every neighborhood of $x$ is disjoint from every neighborhood of $y$.
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What about this? – Juniven Acapulco Mar 21 '17 at 06:29
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It has no proof. – étale-cohomology Mar 21 '17 at 06:30
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1Then see this one. – Juniven Acapulco Mar 21 '17 at 06:31
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What about the 2nd question? – étale-cohomology Mar 21 '17 at 06:50
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1For the "I also wonder" part: that condition never holds in any topological space with at least two points, because the whole space $X$ is a neighbourhood of every point. – Mar 21 '17 at 06:51
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This completely answers my questions. Thanks. – étale-cohomology Mar 21 '17 at 06:58