I am trying to understand when a mode of convergence, i.e., a rule for which sequences in a set $X$ converge and to what value(s?), can give rise to a topology on $X$. There are a number of questions and answers about this subject already on this site, and this answer seems particularly relevant.
Here is what I seem to gather if I understand correctly: A "topological notion of convergence" is one such that the sequential closure
$$ \operatorname{sc}(A) := \{ x \in X | \exists (x_n)_{n\ge1} \subseteq A : x_n \to x \} $$
is a closure operator according to the Kuratowski axioms. This is the case if the mode of convergence satisfies the properties (L1)-(L4) in the linked answer. If you drop assumption (L4), then the sequential closure satisfies all of Kuratowski's axioms except idempotency, so that it still induces a topology but one for which the sequentially closed sets are not necessarily topologically closed.
My problem is that it looks like I can verify the Kuratowski axioms except idempotency while merely assuming (L1) and (L2). Namely:
- We have $\operatorname{sc}(\emptyset) = \emptyset$ since there do not exist any sequences in $\emptyset$
- If $x \in A$, then the constant sequence $(x_n)_{n\ge1} := x$ lies in $A$ and $x_n \to x$ by (L1), so that $x \in \operatorname{sc}(A)$ and thus $A \subseteq \operatorname{sc}(A)$.
- If $x \in \operatorname{sc}(A \cup B)$, then there exists a sequence $(x_n)_{n\ge1} \subseteq A\cup B$ such that $x_n \to x$. Then we can either extract a subsequence in $A$ or in $B$. Indeed, we can try to construct a subsequence in $A$ by letting $n_1$ be the smallest index such that $x_{n_1} \in A$, letting $n_2$ be the next one, etc. If at some point we can no longer find an $n_k$, then $x_n \in B$ for all $n > n_{k-1}$ and the tail of the sequence lies in $B$, which converges to $x$ by (L2) so that $x \in \operatorname{sc}(B) \subseteq \operatorname{sc}(A) \cup \operatorname{sc}(B)$. If we keep finding $n_k$'s, then $(x_{n_k})_{k\ge1}$ is a subsequence of $(x_n)_{n\ge1}$ lying in $A$ with $x_{n_k} \to x$ and thus $x \in \operatorname{sc}(A)$. Conversely, if $x \in \operatorname{sc}(A) \cup \operatorname{sc}(B)$, then there either exists a sequence in $A$ or in $B$ converging to $x$, but in both cases this sequence is also in $A \cup B$.
Am I missing something, either in what I expected to find or in the above argument?