Let $X$ be a topological space and let $A\subseteq X$. Can someone please tell me 1) what the difference between these two expressions is 2) if they are equivalent (if they aren't can someone give me two counterexamples in which in turn each expression holds and the other doesn't?) 3) how one would formulate the first expression in words ?
$ \forall x \in X: \ \ \forall (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$
$ \forall x \in X: \ \ \exists (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$
(The motivation behind this question is that in a course in topology which I took, the first expression was used in the following theorem that stated that $A$ was closed iff the first expresion would hold:
"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$. Then we have: $A$ is closed $\Leftrightarrow$ $\forall (x_n)_n$ in $A,\ x_n \rightarrow x \ \Rightarrow x \in A$ "
And the second expression was used in the following theorem that stated that what the closure of $A$, $c(A)$, would look like:
"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$. Then we have: $c(A)= { x\in X |\ \exists (x_n)_n \textrm{ in }A:\ x_n \rightarrow x } $"
So my guess was, that the above expressions should also be equivalent formally (if we take the st theoretic notation from the second theorem away), since they express the same thing - so I think the proof of the equivalence should not use the fact that our topological space is actually first countable and one of the expression is identical to the fact that $A$ is closed.
Please take note, that this is how the theorems were written down on the blackboard; that is all I have (even if it may be incorrect).
Since I'm not very good with formal notation I may have written the above expression falsely.)