How does convergence of sequences define a topology? Question about the definition of the topology of the Gromov boundary.

Question

I am not sure how the second bullet defines a topology on $\partial X$? What exactly is the topology? What are the open sets, and/or, what are the closed sets?

How does convergence of sequences define a topology?

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Answer 1

If we have a notion of convergence of sequences of $X$ that obeys some minimal axioms, we can define a topology $\mathcal{T}$ by

$O \subseteq X$ is open iff for all $x \in O$, for all sequences $(x_n)$ such that $x_n \to x$ (in our notion of convergence) we know that a tail of the sequence lies in $O$, i.e. : $\exists N \forall n \ge N: x_n \in O$.

$\emptyset$ is open trivially (as there are no $x \in O$ to check the condition for, and $O=X$ as well, of course.

If $O_i, i \in I$ are all open, and $O=\bigcup_i O_i$ then $O$ is open: let $x\in O$ and $x_n \to x$. Then $x \in O_j$ for some $j \in I$ and so a tail of the sequence lies in $O_j$, as $O_j$ is open, hence in $O$.

If $O_1,O_2$ are both open and $x \in O_1 \cap O_2$ and $x_n \to x$, we know that a tail from $N_1$ lies in $O_1$ and a tail from $N_2$ lies in $O_2$, hence the tail from $\max(N_1,N_2)$ lies in $O_1 \cap O_2$, which is thus also open by this definition.

So without using very much of convergence we see that we have defined a topology on $X$ when given a notion of convergence of sequences.

Alternatively, $F$ is closed iff for all $x \in X$ and all sequences $x_n \to x$ where $x_n \in F$ for all $n$, we know that $x \in F$ (so convergent sequences from $F$ can only converge to members of $F$, $F$ is "sequentially closed"): Suppose $F$ is closed, and $x \in X$ and $x_n \to x$ and all $x_n \in F$. If $x \notin F$, $x \in F^\complement$ which is open, so a tail of $(x_n)$ lies in $F^\complement$,contradiction, so $x \in F$. On the other hand, if $F$ is sequentially closed, and if $x \in F^\complement$ and $x_n \to x$. Suppose infinitely many $x_n$ would lie in $F$ then this would define a subsequence $x_{n_k}$ of our sequence inside $F$ and if the convergence obeys the commonly assumed axiom that if a sequence converges to $x$, then so does every subsequence, then we can conclude that $x \in F$ by sequential closedness and we have a contradiction, and we cannot have infinitely many $x_n \in F$ so we have a tail of the sequence inside $F^\complement$, showing that set to be open and $F$ to be closed.

So for a convergence obeying the subsequence axiom we know that sequential closedness and closedness as it follows from the "tail-open" sets is the same. Another standard axiom (that's often trivial but not yet needed here) is that constant sequences converge to their constant value.