Is a closed interval an open set? I read somewhere that a closed interval is not an open set but I don't see why not?
Some definitions in metric space: $(X,d)$
Open Ball: Let $p \in X$ and $r>0$ then $B(p,r) = \{ x \in X : d(x,p) < r\}$
Open set: A $S \subseteq X$ is open set if $\forall p \in S, \exists r>0$ such that $B(p,r) \subseteq S$
Now let's take $X= [a,b]$ and $d(x,y)= |x-y|$ then clearly an open ball can be made for $a<x<b$, for $x= a$, we see $B(a,r) = [a,r) \subseteq X$ if $0<r<b$ and similarly it works for $x=b$, so the closed interval is open set, yeah?

A closed interval is an open set if you consider the relative topology with the interval as the total (this is true for any set of course), since the open sets of the relative topology are the intersections of the open sets of the previous topology with the new total. Better written: Given a set $X$ and a topology $\tau$ defined on it, we define the relative topology on a subset $Y\subseteq X$ as the collection of open sets $\tau_Y:=\{U\cap Y:U\in\tau\}$. In your case $X=\Bbb R$ and $Y=[a,b]$.
This is different of treating a closed interval as a subset of the topological space defined on $\Bbb R$, where your open ball would be $B(a,r)=(a-r,a+r)$, which is clearly not contained in the interval.
