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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 $\{x_{n_k}\}$.

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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 $2^{-n-1}<a<2^{-n}<b<2^{-n+1}$.

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 $0\in\bar C$. $\quad\square$

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  • $\begingroup$ 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. $\endgroup$ – Jeremy Brazas Apr 19 at 12:50
  • $\begingroup$ 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). $\endgroup$ – Jeremy Brazas Apr 19 at 13:04
  • $\begingroup$ @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. $\endgroup$ – Taras Banakh Apr 19 at 14:51
  • $\begingroup$ 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. $\endgroup$ – Jeremy Brazas Apr 19 at 14:56
  • $\begingroup$ I see. Still, it is an intriguing space! $\endgroup$ – Jeremy Brazas Apr 19 at 15:02

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