Sequentially closed $\implies$ closed, but not Fréchet-Urysohn space
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My apologies for the confusion...
I guess I'm asking when a sequential space fails to be an Fréchet-Urysohn space.
general-topologyexamples-counterexamples
asked 2012-04-16
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Arturo: I don't see how that follows logically. " If sequential closure implies closure, then the sequential closure will at least contain the closure". Is there a proof for this? – 2012-04-16
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Arturo: But I'm not sure why seqcl(A) should be closed or even sequentially closed... – 2012-04-16
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I'm not assuming that the seq closure is closed, only that if a set is sequentially closed then it's closed , how do I know that seqcl(A) is sequentially closed? – 2012-04-16
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Then don't use "sequential closure" in your statement, use "sequentially closed". In general, the sequential closure need not be sequentially closed, but I took your statement the way it was written, not the way you now say you meant it. – 2012-04-16
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I've fixed the wording to accurately reflect what you wrote in comment, and deleted my comments as no longer applicable. – 2012-04-16
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The example I've given in [this answer](http://math.stackexchange.com/a/100521/) works too. (Although it was given there for different purpose.) – 2012-04-16
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