# Paths through convergent sequences in $\Delta$-generated spaces

So-called $$\Delta$$-generated spaces are topological spaces in which paths "determine" the topology of the space. In particular, $$X$$ is $$\Delta$$-generated if a set $$U\subseteq X$$ is open (resp. closed) if and only if $$\alpha^{-1}(U)$$ is open (resp. closed) in $$[0,1]$$ for all paths, i.e. continuous functions, $$\alpha:[0,1]\to X$$. The $$\Delta$$-generated spaces form a coreflective convenient category of topological spaces that relate to some familiar properties. For instance, it is true in general that:

first countable and LPC $$\Rightarrow$$ $$\Delta$$-generated $$\Rightarrow$$ sequential and LPC

where LPC abbreviates "locally path connected."

Definition: Let's say that that a topological space $$X$$ is path-sequential if for every convergent sequence $$\{x_n\}\to x$$ in $$X$$, there is a path $$\alpha:[0,1]\to X$$ such that $$\alpha(0)=x$$ and $$\alpha(1/n)=x_n$$ for all $$n\in\mathbb{N}$$.

With some basic arguments, it becomes clear that we have:

first countable and LPC $$\Rightarrow$$ sequential and path-sequential $$\Rightarrow$$ $$\Delta$$-generated $$\Rightarrow$$ sequential and LPC

The first and third implications are definitely not reversible.

Question: Must every $$\Delta$$-generated space be path-sequential?

I am really just interested in the case where $$X$$ is Hausdorff or is at least a US-space, i.e. a space where convergent sequences have unique limits.

Note: it's easy to see that if $$X$$ is $$\Delta$$-generated and a US-space, then for every convergent sequence $$\{x_n\}\to x$$ there exists a path $$\alpha:[0,1]\to X$$ such that $$\alpha(0)=x$$ and $$\alpha(1/k)=x_{n_k}$$ for some subsequence $$\{x_{n_k}\}$$.

The answer to the question is negative. To construct a counterexample, choose a maximal almost disjoint infinite family $$\mathcal A$$ of infinite subsets of $$\omega$$.

Endow $$\mathcal A$$ with the discrete topology and consider the product $$[0,1]\times \mathcal A$$. For every subset $$A\subseteq \omega$$, let $$2^{-A}=\{0\}\cup\{2^{-n}:n\in A\}.$$

Let $$X$$ be the topological sum $$2^{-\omega}\cup([0,1]\times\mathcal A)$$, and $$\sim$$ be the smallest equivalence relation on the space $$X$$ such that $$0\sim (0,A)$$ and $$2^{-n}\sim(2^{-n},A)$$ for every $$A\in\mathcal A$$ and $$n\in A$$. It can be shown that the quotient space $$Y=X/_\sim$$ is a required counterexample: $$Y$$ is $$\Delta$$-generated but not path-sequential (the latter follows from the fact that $$S$$ is not contained in a path-connected compact subspace of $$Y$$).

To be sure that everything works, let us write down the proof of the following

Fact. The space $$Y$$ is $$\Delta$$-generated.

Proof. The space $$Y$$ can be identified with the union $$2^{-\omega}\cup\bigcup_{A\in\mathcal A}([0,1]\setminus 2^{-A})\times\{A\},$$ endowed with a suitable topology. Let $$q:X\to Y$$ be the quotient map.

Take any non-closed set $$C\subset Y$$. If there exists some $$y\in(\bar C \setminus C)\setminus 2^{-\omega}$$, then there exists a unique set $$A\in\mathcal A$$ such that $$y\in ([0,1]\setminus 2^{-A})\times\{A\}$$. In this case for the map $$\gamma_A:[0,1]\to Y$$, $$\gamma_A(t)\mapsto q(t,A)$$, has the desired property: $$\gamma_A^{-1}(C)$$ is not closed in $$[0,1]$$.

So, we assume that $$\bar C\setminus C\subseteq 2^{-\omega}$$. First assume that $$2^{-n}\in\bar C\setminus C$$ for some $$n\in\omega$$. Choose two real numbers $$a,b$$ such that $$2^{-n-1}.

Let $$\mathcal A_n=\{A\in\mathcal A:n\in A\}$$. For every $$A\in\mathcal A_n$$, let $$C_A=C\cap (([a,b]\setminus 2^{-A})\times\{A\})$$. If for some $$A\in\mathcal A_n$$ the set $$C_A$$ contains $$2^{-n}\times\{A\}$$ in its closure, then the map $$\gamma_A:[a,b]\to Y$$, $$\gamma_A:t\mapsto q(t,A)$$, has the required property: the set $$\gamma_A^{-1}(C)$$ is not closed in $$[a,b]$$.

So, assume that for every $$A\in\mathcal A_n$$ the set $$C_A$$ does not contain $$2^{-n}$$ in its closure. By the definition of the quotient topology on $$X$$, the set $$\bigcup_{A\in\mathcal A_n}q((a,b)\setminus \overline C_A)\times\{A\}$$ is an open neighborhood of $$2^{-n}$$ in $$Y$$, which is disjoint with $$C$$. But this contradicts $$2^{-n}\in\overline{C}$$. This contradiction shows that $$\bar C\setminus C=\{0\}$$.

If $$C\cap 2^{-\omega}$$ is infinite, then by the maximality of $$\mathcal A$$, there exists a set $$A\in\mathcal A$$ such that $$C\cap A$$ is infinite. In this case for the map $$\gamma_A:[0,1]\to Y$$, $$\gamma_A:t\mapsto q(t,A)$$, the preimage $$\gamma^{-1}_A(C)\supset C\cap A$$ contains zero in its closure and hence is not closed in $$[0,1]$$.

If the intersection $$C\cap 2^{-\omega}$$ is finite, then we can find a real number $$b\in (0,1]\setminus 2^{-\omega}$$ such that the intersection $$C\cap [0,b]$$ is empty and $$\bar C\cap [0,b]=\{0\}$$. For every $$A\in\mathcal A$$ consider the set $$C_A=C\cap ([0,b]\setminus 2^{-A})\times\{A\}$$. If for some $$A\in\mathcal A$$ the set $$C_A$$ contains zero in its closure in $$[0,b]$$, then for the map $$\gamma_A:[0,1]\to Y$$, $$\gamma_A:t\mapsto q(t,A)$$, the preimage $$\gamma_A^{-1}(C)=C_A$$ contains zero in its closure and hence is not closed in $$[0,1]$$.

So, we assume that for every $$A\in\overline A$$ the closure $$\overline{C_A}$$ does not contain zero. Since $$\overline{C_A}\subset \overline C$$ and $$\overline C\cap [0,b]=\{0\}$$, the set $$[0,b)\cup\bigcup_{A\in\mathcal A}(([0,b)\setminus 2^{-A})\setminus \overline C_A)\times\{A\}$$ is an open neighborhood of zero, which is disjoint with the set $$C$$. But this contradicts $$0\in\bar C$$. $$\quad\square$$

• This is an interesting construction. I am trying to sort out what properties of a maximal almost disjoint infinite family of subsets of $\omega$ would really be used here. – Jeremy Brazas Apr 19 at 12:50
• Also, I'm having a little trouble seeing why $Y$ is even locally path-connected at the image of $0$ (in the given quotient topology). – Jeremy Brazas Apr 19 at 13:04
• @JeremyBrazas It nice that you finally commented on this construction. The maximality of the almost disjoint family is necessary for killing all possible convergent subsequences. – Taras Banakh Apr 19 at 14:51
• Yes, a space is $\Delta$-generated iff it is a quotient of a disjoint union of copies of $[0,1]$. Since quotients of LPC spaces are LPC, every $\Delta$-generated space must be LPC. – Jeremy Brazas Apr 19 at 14:56
• I see. Still, it is an intriguing space! – Jeremy Brazas Apr 19 at 15:02