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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 $X$.

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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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 $*.$


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 $*.$

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  • $\begingroup$ Thank you for the answer, but your third condition contradicts the second one. In fact, the second is true but the third is not. $\endgroup$ – Taras Banakh Jan 29 at 12:23
  • $\begingroup$ @TarasBanakh: Sorry, I hadn't thought that through. I've now come up with a different construction using CH. $\endgroup$ – Dap Feb 1 at 7:11
  • $\begingroup$ Thank you for another answer. I have some remarks: (1) it seems that the sequences $(S_\alpha)$ and $(f_\alpha)$ can be constructed under the assumtion $\mathfrak p=\mathfrak c$ (which is weaker than CH and follows from MA, actually it is equivalently to some version of MA); (2) usually by $\kappa^+$ the successor cardinal to $\kappa$ is denoted. So, your $\omega^+$ is a bit misleading; your can write $\omega+1$ or $\bar\omega$ instead. $\endgroup$ – Taras Banakh Feb 1 at 10:17
  • $\begingroup$ And the main problem is with your definition of the topology at $*$: as $\alpha$ icreases so do the sets $U_{n,\alpha}$ but it should be vice-versa. If you replace $\le$ by $\ge$ in the definition of $U_{n,\alpha}$, then you will obtain the standard compact metrizable topology on $(\omega\times\{\omega\})\cup\{*\}$ and this will not allow you to prove that your space is not sequential. $\endgroup$ – Taras Banakh Feb 1 at 10:22
  • $\begingroup$ @TarasBanakh: Thanks for the comments! The sets $U_{n,\alpha}$ are not increasing in $\alpha$ - they just form a subbase of $*.$ The sequence $S_\alpha$ is $\supseteq^*$-increasing i.e. decreasing modulo finite sets. I've fixed the ordinal notation. $\endgroup$ – Dap Feb 2 at 7:12
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$2^\kappa$ for uncountable $\kappa$ is a ZFC example of an almost sequential non-sequential space.

The easiest way to see why it's not sequential is to note that there is a set $A \subset 2^\kappa$ and a point $p \in \overline{A}$ such that $p$ is not in the closure of any countable subset of $A$. It suffices to take $A=\{x \in 2^\kappa: |x^{-1}(1)| < \aleph_0 \}$ and $p \in 2^\kappa$ to be the function constantly equal to 1.

To see why it's almost sequential, let $x$ be any point in $2^\kappa$ and let $D=\{y \in 2^\kappa: |\{\alpha < \kappa: x(\alpha) \neq y(\alpha) \}| < \aleph_0 \}$. Then $D$ is a sequential (even Fréchet-Urysohn) dense subspace of $2^\kappa$ which contains $x$.

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  • $\begingroup$ Thank you! It is so evident! $\endgroup$ – Taras Banakh Feb 6 at 10:50

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