# 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I've fixed the wording to accurately reflect what you wrote in comment, and deleted my comments as no longer applicable.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
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## 2 Answers

A sequential space is Fréchet if and only if it does not contain a copy of the Arens space. This theorem is proved, and the Arens space is defined, in Dan Ma’s Topology Blog. (Note the links at the end of the article to previous posts on sequential spaces that may also be of interest.) Note that the Arens space is not the Arens-Fort space described in Wikipedia; the latter is a non-sequential subspace of the former.

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The standard example for a sequential space that is not Frechet is Arens space. The space is described in a general topology book or in wikipedia.

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