Closed subset:= If $(a_n)_{n\in\mathbb{N}}$ converges and the elements of the sequence are all in $A$ then the Limit sits in $A$.
If I could Show that there exists a convergent sequence where all the Elements of the sequence are in $A$ and the Limit is not in $A$ would it be equivalent to say that $A$ is not a closed subset?$(*)$
I.e $A$ not a closed subset $\iff \exists (a_n)_n\in\mathbb{N},a_n\in A \forall_{n\in\mathbb{N}}$ convergent and $lim_{n\rightarrow\infty}a_n\notin A$
I have the Feeling that something is wrong because the Definition says:
$A$ closed subset $\iff \forall(a_n)_{n\in\mathbb{N}},a_n\in A \forall_{n\in\mathbb{N}} convergent\Rightarrow lim_{n\rightarrow\infty}a_n\in A$
Therefor the Negation would have to be
$A$ not closed subset $\iff \forall(a_n)_n\in\mathbb{N},a_n\in A \forall_{n\in\mathbb{N}}$ convergent and $lim_{n\rightarrow\infty}a_n\notin A$
If I have $(*)$ can it help me to prove:
$A$ closed subset $\iff A^C$ is open
If you define closed as sequentially closed (so every sequence in $A$ that converges in $X$ to $p$ has $ p \in A$ too), then the logical negation of that
$A$ is not closed iff there exists some sequence $(a_n)$ converging to $p$ where all $a_n \in A$ but $p \notin A$. (so one such sequence is enough to disprove closedness).
I don't see how that helps you proving $A$ closed iff its complement is open if you don't also have a definition of open sets in terms of sequences.
A final word of caution: said equivalence of closed and sequentially closed does not holds in all spaces (but it does in e.g. metric spaces, a very common case).