Intuition behind closed subsets of a metric space? Reading for my exam in real analysis, I struggle with the definition of a closed subset of a metric space.
Consider a metric space $$(X,d)$$
Then consider a subset of this space$$F$$
What the book is trying to convince me of is that F is closed iff a convergent sequence in F always has a limit belonging to F.
The intuition that:
$$x_1,\dots,x_n,\dots\in F$$
$${x_n}\rightarrow a $$
where
$$a\in F$$
for all convergent sequences in F implies that F is closed makes sense to me.
What I am struggling with is understanding the converse. What is it about F containing all of its boundary points that makes it impossible to have a sequence in F converging to some limit outside of F (even though the sequence will never get there)?
Edit:
Is the last sentence in the parenthesis actually the reason I am looking for? That it is impossible to get arbitrarily close to some point outside of F, because when you get really close, then you're not necessarily in F at all?
 A: I'm slightly confused about what you think the converse is, and what your definition of a closed set is. You're happy that $$\text{all convergent sequences in $F$ converge to a limit in }F \implies F \text{ is closed}$$ so what you're missing is $$ F \text{ is closed}\implies\text{all convergent sequences in $F$ converge to a limit in }F $$
But "$F$ consisting only of interior points" is equivalent to $F$ being open, which doesn't have anything to do with this statement. Since $F$ closed $\iff$ the complement of $F$ is open, one can argue the following if you want to think about interior points:

Suppose a sequence converges to some point $x\not\in F$. But the complement of $F$ is open, so $x$ is interior to the complement, and there is an open ball $B_x$ which also does not intersect $F$. But then the sequence cannot lie in $F$ since it must eventually lie entirely within $B_x$ since it tends to $x$. Intuitively: there's a neighbourhood of the limit which is stuck outside $F$.

Edit: If instead you want to think in terms of $F$ containing all 'boundary' points or limit points, then the result you want is true by definition! 'Closed' literally means 'closed under taking limits'.
