# Sequentially closed $\implies$ closed, but not Fréchet-Urysohn space

My apologies for the confusion...

I guess I'm asking when a sequential space fails to be an Fréchet-Urysohn space.

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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? – hello Apr 16 '12 at 3:19
Arturo: But I'm not sure why seqcl(A) should be closed or even sequentially closed... – hello Apr 16 '12 at 3:37
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? – hello Apr 16 '12 at 3:41
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. – Arturo Magidin Apr 16 '12 at 3:42
I've fixed the wording to accurately reflect what you wrote in comment, and deleted my comments as no longer applicable. – Arturo Magidin Apr 16 '12 at 3:53