Re: Hausdorff, not-first countable space where sequential closure = closure?

On Fri, 30 Dec 2005 Rufus.Zee@xxxxxxxxx wrote:

> > Yes, R with topology { {r}, A | r /= 0, 0 in A, R\A finite }
>
> In other words, every non-zero point is isolated, and every
> neighborhood of 0 is cofinite.
>
> So this space fails to have a countable basis at 0, is clearly
> Hausdorff. The closure of any finite set is itself; the closure of any
> infinite set is itself union {0}. Any infinite set admits a sequence
> converging to 0. So this works; it is even normal, since disjoint
> closed sets are disjoint finite set are disjoint open sets containing
> themselves. It is even compact.
>
> Any idea what conditions in addition to the "sequential closure =
> closure" would imply first countability? Compact Hausdorff evidently
> doesn't do the trick; perhaps continuums? (using: A continuum is a
> connected compact Hausdorff space.)
>
2nd countable ==> 1st countable

> In summary: Let us say a space is a 'sequence lemma space' or SL space
> if the notions of closure and sequential closure coincide. An example
> has been given of a compact Hausdorff SL space that is not first
> countable. Does there exist an SL continuum that is not first
> countable? And is there a standard name for what I'm calling an SL
> space?
>
Perhaps [0,1]^[0,1]. No, as far as I know, it's all yours.
.

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