X is a topology; p $\notin$ A $\subset$ X;
if p is a limit point of A; that is, every [open subset of X containing p] has non-empty intersection with A,
then does there always exist a sequence of A that converges to p?
this is true if X is a metric space, however i am not sure whether this is true or not in general cases.
if it's false, please give me a counterexample.
if this is true, does the proof require any sort of choice-related axioms other than ZFC's?
one more thing;
f : X $\rightarrow$ Y, where X and Y are topologies; x $\in$ X;
assume that for every [sequence x$_n$ of X that converges to x], f(x$_n$) converges to f(x).
then is f continuous on x?; that is, for every [open subset U $\subset$ Y, containing f(x)] there exists an open subset V $\subset$ X s.t. f(V) $\subset$ U
if X is a metric space, one can prove the above by contradiction, making a sequence that converges to x but its image not converging to f(x).
is this true in general?
It's true in a metric space, as a metric space is first countable and we can use countable choice to show it (well within ZFC).
It's not true in general: let $X$ be the reals (or any uncountable set) in the co-countable topology. Any sequence in $X$ that converges to $p$ is equal to $p$ after finitely many terms (eventually constant); many proofs exist for this on this site.
So $0 \in \overline{A}$ where $A= (0,\infty)$ but no sequence in $A$ converges to $0$. Also, the identity map from $X$ to $Y$, the reals in the usual topology, is not continuous but trivially sequentially continuous.