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Can anyone think of an example of a topological space that admits sequentially open sets that are not open?

A subset $U\subseteq X$ is called sequentially open if the following is true: Whenever some sequence $x_n$ in $X$ is convergent to $x\in U$, then there is $n_0$ such that $x_n\in U$ for each $n\ge n_0$. (I.e., the sequence $x_n$ is eventually in $U$.)

An equivalent condition is that the complement is sequentially closed. That means that for any convergent sequence which lies entirely in the complement $X\setminus U$, the limit also belongs to $X\setminus U$.

Thanks!

Martin Sleziak
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Mark
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1 Answers1

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Let $X$ be an uncountable set with the co-countable topology; a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converges to a point $x\in X$ if and only if there is an $m\in\Bbb N$ such that $x_n=x$ for all $n\ge m$. Let $A$ be an uncountable subset of $X$ with uncountable complement; then $A$ is not open, but any sequence converging to a point of $A$ is eventually in $A$.

Of course that space is only $T_1$, but we can do better. Let $\alpha$ be any ordinal greater than $\omega_1$, and let $X$ be $\alpha$ with the order topology. $X$ is quite nice, being hereditarily normal. Let $A=\{\xi\in X:\omega_1\le\xi\}$; $A$ is not open in $X$, but $A$ is sequentially open, since every sequence frequently in $X\setminus A$ either converges to a point of $X\setminus A$ or fails to converge.

Brian M. Scott
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  • I didn't understand the second example, and in particular how show sequential openness. Per the definition of the OP, shouldn't we show that for every sequence which lies entirely in X\A the limit belongs to X\A. Why is it okay that the sequence is only frequently in X\A and that it may not converge? – roi_saumon Dec 05 '20 at 15:48
  • @roi_saumon: We’re not showing that $X\setminus A$ is sequentially open: we’re showing that $A$ is. To do that, we need to show that if a sequence converges to some point of $A$, then it is eventually in $A$. Equivalently, we can show that if a sequence is not eventually in $A$, then it does not converge to a point of $A$. If a sequence is not eventually in $A$, then it is frequently in $X\setminus A$, and in that case it either converges to a point of $X\setminus A$ or fails to converge at all, so it definitely doesn’t converge to a point of $A$. – Brian M. Scott Dec 05 '20 at 18:13
  • Thank you I think I understand id now. Just to be sure : If $ (x_n){n\ge 1}$ is a sequence frequently in $X\setminus A$ that would converge to an element $x\in A$, then every subsequence converges to $x$. But since $x_n$ is frequently in $X\setminus A$ then there is a subsequence $(x{n_k})k\subset X\setminus A$ that converges to $x$, which is not possible because the $x{n_k}$ are countable and there limit $\cup_k x_{n_k}$ would be countable and but $x$ is uncountable. – roi_saumon Dec 06 '20 at 00:50
  • @roi_saumon: Yes, that’s correct. – Brian M. Scott Dec 06 '20 at 00:52
  • thank you. In the first example, shouldn't it be the co-countable topology that has this property for converging sequences? – roi_saumon Dec 06 '20 at 01:30
  • @roi_saumon: Ack! I can’t believe that it took almost two months for someone to catch that error. Many thanks; I've fixed it now. – Brian M. Scott Dec 06 '20 at 01:35