What criteria must a family of sequences meet in order to induce a topology?

Question

What criteria must a family of sequences meet in order to induce a topology?

An obvious condition is that constant sequences should converge, but is there more?

Also, I feel like this is a duplicate, but has been oddly difficult to Google...

Answer 1

Following Engelking General Topology ex 1.7.18/19, that follow papers by Fréchet and Urysohn and Kisyński, look up the references there.) Let $X^\mathbb{N}$ be the set of all sequences on $X$.

Suppose we have a function from $\lambda: \mathscr{C} \subseteq X^\mathbb{N} \to X$, where $\mathscr{C}$ is a designated set of "convergent sequences", and $\lambda((x_n)_n) = x$ says that $x$ is the limit of the sequence, also denoted $x_n \to x \,(n \to \infty)$.

Then $(X,\lambda)$ is called a $\mathscr{L}^\ast$-space when it satisfies the following three axioms:

L1 If $x_n = x$ for all $n$, then $x_n \to x$. (constant sequences converge to their constant.)

L2 If $x_n \to x$ then $x_{n_k} \to x \,(k \to \infty)$ for every subsequence of $x_n$ (or $\lambda((x_{n_k})_k) = x$)

L3 If a sequence $(x_n)_n$ does not converge to $x$, then there is a subsequence $(x_{n_k})_k$ of $(x_n)_n$ such that no subsequence of $(x_{n_k})_k$ converges to $x$.

An alternative logical formulation of L3 is: if every subsequence of a sequence has a sub-subsequence that converges to $x$, then the original sequence converges to $x$.

Then we can define a so-called closure operator on $X$ by:

$x \in \overline{A}$ iff $\exists (x_n)_n \in \mathscr{C} \cap A^\mathbb{N}: \lambda((x_n)_n) =x$

so $x$ is in the closure of $A$ iff there is a sequence from $A$ converging to $x$.

This satisfies the first 3 axioms of a closure operator but not idempotence $\overline{\overline{A}} = \overline{A}$, we only have a preclosure which does define a (the so-called Fréchet) topology (via "$A$ is closed" iff $A = \overline{A}$), but such that the closure-operator in this induced topology is not always equal to the orginal closure operation from the convergence, which is maybe not what you'd like.

In order to get that extra condition we need we define another axiom

L4 If $x_n \to x (n \to \infty)$ and for every $n$, $x^n_k \to x_n (k \to \infty)$ then there exist integer sequences $n_1, n_2, \ldots$ ,$k_1, k_2, \ldots$ such that $x^{n_l}_{k_l} \to x \,(l \to \infty)$

And we can show that $\overline{A}$ defined above from the convergence is idempotent iff it also satisfies L4. A space satisfying L1-L4 is called an $\mathscr{S}^\ast$ space, and in those the sequential closure equals its topological closure and moreover the topological convergence is the same as the formal convergence relation $\lambda$ we started with.

Kisyński (1960) then showed that in an $\mathscr{L}^\ast$ space we can also define a topology directly, without a preclosure operation:

$C \subseteq X$ is closed if for every sequence from $C$ that converges, its limit lies in $C$ as well.

This is the sequential topology induced by $\lambda$ (as opposed to the Fréchet topology induced by the preclosure) and it can be shown that an $\mathscr{L}^\ast$ topology is $T_1$ in this sequential topology. In an $\mathscr{S}^\ast$ space the Fréchet topology and the sequential topology coincide.

Exercise 1.7.20 then goes on to connect these ideas to the notions of a sequential and a Fréchet topological space. I warmly commend this great resource to your attention. You could go back to the original papers if you read French.