Sequentially open sets but not open

Question

Can anyone think of an example of a topological space that admits sequentially open sets that are not open?
A subset $Usubseteq X$ is called sequentially open if the following is true: Whenever some sequence $x_n$ in $X$ is convergent to $xin U$, then there is $n_0$ such that $x_nin U$ for each $nge n_0$. (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 $Xsetminus U$, the limit also belongs to $Xsetminus U$.
Thanks!

Brian M. Scott · Accepted Answer

Let $X$ be an uncountable set with the co-countable topology; a sequence $langle x_n:ninBbb Nrangle$ in $X$ converges to a point $xin X$ if and only if there is an $minBbb N$ such that $x_n=x$ for all $nge m$. 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 $A={xiin X:omega_1lexi}$; $A$ is not open in $X$, but $A$ is sequentially open, since every sequence frequently in $Xsetminus A$ either converges to a point of $Xsetminus A$ or fails to converge.

Sequentially open sets but not open

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