# Are almost sequential spaces sequential?

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)?

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 $$*.$$

• Thank you for the answer, but your third condition contradicts the second one. In fact, the second is true but the third is not. – Taras Banakh Jan 29 at 12:23
• @TarasBanakh: Sorry, I hadn't thought that through. I've now come up with a different construction using CH. – Dap Feb 1 at 7:11
• 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. – Taras Banakh Feb 1 at 10:17
• 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. – Taras Banakh Feb 1 at 10:22
• @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. – Dap Feb 2 at 7:12

$$2^\kappa$$ 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 $$x$$.

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