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A topology can be characterized by net convergence generally.

However, none of the counterexamples I have learnt where sequence convergence does not characterize a topology is Hausdorff.

Such as cocountable topology on an uncountable space, where every set is closed under sequence convergence.

I believe there are Hausdorff counterexamples but can't find one.

Yuz
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1 Answers1

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A good source of counterexamples are the ordinal numbers with the order topology, which is Hausdorff.

The first uncountable ordinal $\omega_1$ is what you're looking for.

Consider a sequence $(\alpha_n)$ in $\omega_1$. Then each term of the sequence is a countable ordinal and the supremum of a countable set of countable ordinals is countable. Therefore the sequence is inside $\beta+1$, where $\beta=\sup\{\alpha_n\}$. The ordinal $\beta+1$ is, like all successor ordinals, compact.

Hence the sequence has a convergent subsequence.

On the other hand, $\omega_1$ is not compact, because a limit ordinal is not compact.

More simply, as suggested by Henno Brandsma, the subspace $\omega_1$ of $\omega_1+1$ is dense, but $\omega_1$ is not a limit of a sequence in $\omega_1$ (same argument as before).

Henno Brandsma
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egreg
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    $\omega_1 + 1 = [0,\omega_1]$ has a maximum that is in the closure of $[0,\omega_1)$ but has no convergent sequence converging to it. Both these spaces are hereditarily normal (so much better than Hausdorff). – Henno Brandsma Dec 12 '20 at 10:03
  • @HennoBrandsma I added your suggestion, thanks. – egreg Dec 12 '20 at 10:13
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    @Yuz - Also consider the long line, which is $[0, \omega_1) \times [0,1)$ (the real interval) under the lexigraphic order. It also satisfies sequentially closed $\not\Rightarrow$ closed for the exact same reason but is also path connected and locally euclidean, and has a number of other properties that are useful for further counter-examples. – Paul Sinclair Dec 12 '20 at 21:21