Let (X,d) be a metric space and assme that A$\displaystyle \subset$X. Prove that $\displaystyle x \in Cl(A)$ iff there exists a sequence in A converging to x.
Here's a rushed proof. you can streamline it.
Assume there is a sequence of points in $\displaystyle A$ that converges to $\displaystyle x$. Then $\displaystyle x$ lies in any closed set containing $\displaystyle A$. That is, $\displaystyle x \in \text{Cl}(A)$.
Conversely, assume $\displaystyle x \in \text{Cl}(A)$. Then the "ball" of radius $\displaystyle 1/k$ centered at $\displaystyle x$ will contain points in $\displaystyle A$. For each $\displaystyle k$, pick one of these points and call it $\displaystyle a_k$, then the sequence $\displaystyle \{ a_k \}$ converges to $\displaystyle x$
I assume you know how to describe a "ball" in a metric space. we derive this from the metric
gee, i don't know of a theorem that says we can do this. it just makes sense that we can based on our definitions.
what does it mean to be in $\displaystyle \text{Cl}(A)$? Well, it means that if $\displaystyle x \in \text{Cl}(A)$ then any "open ball" centered at $\displaystyle x$ will contain points in $\displaystyle A$. what does "open ball" mean in a metric space? the "open ball" centered at x is the set $\displaystyle \{ y \in A \mid d(x,y) < r , \text{ for some given } r > 0 \}$. so being in the closure means that this set is nonempty. if the set is nonempty, what's stopping us from picking elements in the set?