just a quick question for my own clarification.
Consider the metric space $(X,d)$ with $A \subset X$.
we know $A \subset_{closed} X$ if $A$ is finite, or $\forall$ infinite sequence $x_n \in A ~ \exists ~x \in A$ such that $x_n \rightarrow x$ then for visualisation purposes does this mean that the only instance (putting open & closed sets to the side for the moment) that $A \subset_{open} X$ is if there exists a divergent infinite sequence in A?
i know A is open if $Int(A) = A$ (also the requirements for $x \in Int(A)$) and this implies $X \setminus A$ is closed. Further the above suggests that $X \setminus A$ can be either finite or infinite, but surely $Int(A)$ must be infinite?
anyone able to give some wise words which might make sense of this?
thanks for taking the time to answer.
Your definition of closed is wrong (you're suggesting that all sequences from $A$ actually converge). If you want a sequence-based definition:
$A$ is closed iff for all sequences $(x_n)$ in $X$ (so these are just functions from $\mathbb{N}$ to $X$, they could be constant, or whatever you please) such that $x_n \in A$ for all $n$ and if $x_n \to x$ (in $X$), then $x \in A$. So it says nothing about there being any non-trivial convergent sequences in $A$, but that if they exist their limit must belong to $A$ as well: informally, you cannot converge to a non-$A$ point with just sequence points from $A$.
This works in metric spaces and general spaces "like it" (so-called sequential spaces).
Being open has a comparable equivalent definition (in metric or similar spaces) in terms of sequences as well: $O$ is open iff for all $x \in O$, and if a sequence $(x_n)$ in $X$ converges to $x$, all but at most finitely many $x_n$ must lie in $O$ as well (i.e. the set $\{n: x_n \notin O\}$ is a finite subset of $\mathbb{N}$). No info on divergent sequences, but a conditional statement about sequences that do converge, just as for closed sets.