A topological space $X$ is called


$\bullet$ *sequential* if for each non-closed subset $A\subset X$ there exists a sequence $\{a\_n\}\_{n\in\omega}\subset A$ that converges to a point $a\notin A$;


$\bullet$ *almost sequential* if each point $x\in X$ is contained in a dense sequential subspace of $X$.



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> **Question.** Is there an almost sequential regular space $X$ which is not sequential (and moreover, contains a closed countable subspace $F\subset X$ that has no non-trivial convergent sequences)?
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Are almost sequential spaces sequential?
I believe there is a regular non-sequential almost sequential space in ZFC+CH. (See below for a construction using a weaker assumption.)


For $S,T\subseteq \omega$ let $S\subseteq^\* T$ denote inclusion modulo finite sets i.e. $S\setminus T$ is finite.
For $f,g:\omega\to\omega$ let $f\leq^\* g$ denote dominance modulo finite sets i.e. $f(n)\leq g(n)$ except for finitely many $n.$
Greek letters will denote elements of $\omega\_1.$


The construction will make use of an ultrafilter $\mathcal U$ on $\omega,$
a cofinal increasing $\omega\_1$-sequence $S\_\alpha$ in $(\mathcal U,\supseteq^\*),$
and a cofinal increasing $\omega\_1$ sequence $f\_\alpha$ in $(\omega^\omega,\leq^\*).$
So $\alpha<\beta$ implies $S\_\alpha\supset^\* S\_\beta,$ and for every set $S$ there is $\alpha$ such that $S\supseteq^\* S\_\alpha$
or $\omega\setminus S\supseteq^\* S\_\alpha.$ And $\alpha<\beta$ also implies $f\_\alpha\leq^\*f\_\beta,$ and for every $f$ there is
$\alpha$ such that $f\leq^\* f\_\alpha.$
I will also require $S\_0=\omega.$
These are easy to construct under CH by transfinite induction.
(Specifically, we can take $S\_\alpha$ to be a strictly $\subseteq^\*$-decreasing subsequence of the sets called $X\_\alpha$
in the construction of a Ramsey ultrafilter in Jech's *Set theory* Theorem 7.8 (3rd ed). $f\_\alpha(n)$ can be constructed by a very similar argument.)


The ordinals $\omega$ and $\omega+1$ have the ordinal topology.
Let $X$ be the topological space on the set $(\omega\times(\omega+1))\cup\{\*\}$ generated by open sets in $\omega\times(\omega+1)$
and the sets $U\_{\alpha,n}$ defined for all $\alpha\in\omega\_1$ and all integers $n$ by
$$U\_{\alpha,n}=\{\*\}\cup\{(x,y)\mid x>n\text{ and either }x\in S\_\alpha\text{ or }y\leq f\_\alpha(x)\}.$$
The sets $U\_{\alpha,n}$ are a neighborhood subbase of $\{\*\}.$


**$X$ is regular.** It has a subbase of clopen sets.


**$X$ is not sequential.** The subspace $A=\omega\times\{\omega\}$ has $\{\*\}$ as a limit point.
This is because any finite intersection $U\_{\alpha\_1,n\_1}\cap \dots\cap U\_{\alpha\_k,n\_k}\cap A$
is just $S\_{\max(\alpha\_i)}\times\{\omega\}$ minus a finite set, and is therefore non-empty.
Suppose for contradiction that a sequence $(x\_n,\omega)$ converges to $\*.$ Split $\{x\_n\}$ into two infinite sets.
One of these sets, call it $S,$ is not in $\mathcal U.$ There is $\alpha$ such that $\omega\setminus S\supseteq^\* S\_\alpha.$
The corresponding subsequence therefore lies outside $U\_{\alpha,n}$ for sufficiently large $n.$


**$X$ is almost sequential.** Each $x\neq \*$ lies in the dense sequential subspace $X\setminus\{\*\}.$
The point $\*$ lies in the subspace $A=(\omega\times\omega)\cup \{\*\}$ which is clearly dense, and it remains to show that it is sequential. The only problem is at $\*.$ Accordingly, consider a set $C\subseteq A\setminus\{\*\}$ such that $\*$ is a limit point of $C,$
and we need to exhibit a sequence converging to $\*.$
Define $C\_\gamma=\{(x,y)\in C\mid x\in S\_\gamma\}.$


First consider the case that $\*$ is a limit point of $C\_\gamma$ for every $\gamma.$
Let $D=\{x\mid \exists y.(x,y)\in C\}.$
Pick any function $f:\omega\to\omega$ with $(x,f(x))\in C$ for each $x\in D.$
Pick $\alpha$ such that $f\leq^\* f\_\alpha.$ Take any strictly increasing sequence $x\_n$ in $D\cap S\_\alpha.$ Consider an arbitrary $U\_{\beta,N}.$
We either have $\beta\leq\alpha$ giving $x\_n\in S\_\beta$ eventually, or $\beta\geq\alpha$ giving $f(x\_n)\leq f\_\beta(x\_n)$ eventually.
This proves that $(x\_n,f(x\_n))$ converges to $\*.$


Now assume that $\*$ is not a limit point of $C\_\gamma,$ for some $\gamma.$
Take $\gamma$ to be minimal. So $\*$ is a limit point of $C\_{\beta}$ if and only if $\beta<\gamma.$
Since $0\neq\gamma<\omega\_1$ there is a countable sequence $\beta\_n$ with $\sup(\beta\_n+1)=\gamma.$
For each $n,N$ the set $C\_{\beta\_n}\setminus C\_\gamma$ must intersect the neighborhood $U\_{\gamma,N},$
which means there are $(x,y)\in C\_{\beta\_n}\setminus C\_\gamma$ with $x>N$ and $y\leq f\_\gamma(x).$
Therefore $C\_{\beta\_1}\cap\dots\cap C\_{\beta\_n}\setminus C\_\gamma$ (which might be smaller than $C\_{\beta\_n}\setminus C\_\gamma$ at a finite number of $x$-coordinates) contains some $(x\_n,y\_n)$ with $x\_n>N$ and $y\_n\leq f\_\gamma(x\_n).$ We can pick such a choice of $(x\_n,y\_n)$ for each $n,$ using $N$ to ensure that $x\_n$ is strictly increasing.
This construction ensures that for each $\beta<\gamma$ the sequence $(x\_n,y\_n)$ eventually lies in $C\_{\beta},$
and $y\_n\leq f\_\gamma(x\_n)$ for all $n.$ So for each $\beta,N$ the sequence $(x\_n,y\_n)$ eventually lies in $U\_{\beta,N}.$
This proves that $(x\_n,y\_n)$ converges to $\*.$




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If I understand correctly, the above argument relied on $\mathfrak{u}=\mathfrak{d}=\aleph\_1.$ I believe this assumption can be weakened to $\mathfrak{d}=\aleph\_1,$ which is used for the diagonalization at the end.


The argument is very similar. We still have $f\_\alpha$ but no ultrafilter nor $S\_\alpha.$ I will assume $f\_0(n)=0.$ Pick a bijective function $p:\omega\times\omega\to\omega.$ The base set for $X$ will instead be $(\omega\times\omega\times(\omega+1))\cup\{\*\},$ and $U\_{\alpha,N}$ will instead be
$$U\_{\alpha,N}=\{\*\}\cup\{(x,y,z)\mid x>N\text{ and either }y\geq f\_\alpha(x) \text{ or }z\leq f\_\alpha(p(x,y))\}.$$


This $X$ also has a subbase of clopen sets, and is not sequential because it has a non-sequential closed subspace, the Arens-Fort space $(\omega\times\omega\times\{\omega\})\cup\{\*\}.$
As before, $X\setminus\{\*\}$ is sequential.
To show that $(\omega\times\omega\times\omega)\cup\{\*\}$ is sequential,
consider a set $C\subseteq \omega\times\omega\times\omega$ such that $\*$ is a limit point of $C.$ Define $C\_\gamma=\{(x,y,z)\in C\mid y\geq f\_\gamma(x)\}.$


First consider the case that $\*$ is a limit point of $C\_\gamma$ for every $\gamma.$
Let $D=\{(x,y)\mid \exists z.(x,y,z)\in C\}.$
Pick any function $f:\omega\to\omega$ with $(x,y,f(p(x,y)))\in C$ for each $(x,y)\in D.$
Pick $\alpha$ such that $f\leq^\* f\_\alpha.$ Pick a sequence of pairs $(x\_n,y\_n)\in D$ with $x\_n\to\infty$ and $y\_n\geq f\_\alpha(x\_n)$ (these exist because $C\_\alpha$ is not bounded in the $x$ direction). Consider an arbitrary $U\_{\beta,N}.$
We either have $\beta\leq\alpha$ giving $y\_n\geq f\_\beta(x\_n)$ eventually, or $\beta\geq\alpha$ giving $f(p(x\_n,y\_n))\leq f\_\beta(p(x\_n,y\_n))$ eventually.
This proves that $(x\_n,y\_n,f(p(x\_n,y\_n)))$ converges to $\*.$


Now assume that $\*$ is not a limit point of $C\_\gamma,$ for some $\gamma.$
Take $\gamma$ to be minimal. So $\*$ is a limit point of $C\_{\beta}$ if and only if $\beta<\gamma.$
Since $0\neq\gamma<\omega\_1$ there is a countable sequence $\beta\_n$ with $\sup(\beta\_n+1)=\gamma.$
For each $n,N$ the set $C\_{\beta\_n}\setminus C\_\gamma$ must intersect the neighborhood $U\_{\gamma,N},$
which means there are $(x,y,z)\in C\_{\beta\_n}\setminus C\_\gamma$ with $x>N$ and $z\leq f\_\gamma(p(x,y)).$
Therefore $C\_{\beta\_1}\cap\dots\cap C\_{\beta\_n}\setminus C\_\gamma$ (which might be smaller than $C\_{\beta\_n}\setminus C\_\gamma$ at a finite number of $x$-coordinates) contains some $(x\_n,y\_n,z\_n)$ with $x\_n>N$ and $z\_n\leq f\_\gamma(p(x\_n,y\_n)).$ We can pick such a choice of $(x\_n,y\_n,z\_n)$ for each $n,$ using $N$ to ensure that $x\_n$ is strictly increasing.
This construction ensures that for each $\beta<\gamma$ the sequence $(x\_n,y\_n,z\_n)$ eventually lies in $C\_{\beta},$
and $z\_n\leq f\_\gamma(p(x\_n,y\_n))$ for all $n.$ So for each $\beta,N$ the sequence $(x\_n,y\_n,z\_n)$ eventually lies in $U\_{\beta,N}.$
This proves that $(x\_n,y\_n,z\_n)$ converges to $\*.$


