Confusion about closed sets

Question

I'm currently learning about open and closed sets (in the context of compactness) and I've run into something I don't quite understand. One of the definitions of a closed set is that its complement is open. However any closed interval $[a,b]$ (denote as $A$ for simplicity) is considered to be a closed set, but its complement $(-\infty,a)\cup(b,\infty)$ (denoted $A^c)$ doesn't appear to be open to me by the definition that every element of an open set has a neighbourhood contained in the open set. So to me, picking some value close to $a$ or $b$, lets call it $c$, $\exists$ $\epsilon>0$, s.t. the set $(c-\epsilon,c+\epsilon)$ will not entirely be contained in $A^c$ (some of it will be within the pocket $[a,b]$ which is not part of $A^c)$. I'm not sure if I'm misunderstanding some of these definitions, or maybe I'm misusing them completely but any clarity would be greatly appreciated.

Answer 1

If some positive $c$ is in the complement of $[a,b]$ then $c > b$. If you let $\epsilon = (c-b)/2$ then the $\epsilon$-neighborhood of $c$ is in the complement of $[a,b]$.

Answer 2

To fix the ideas, take $a=0$ and $b=1$. Then $$ A^c = (-\infty,0) \cup (1,\infty). $$ Take $c \in A^c$, it is either in $(-\infty,0)$ or in $(1,\infty)$. To fix the ideas, assume it is in $(1,\infty)$. Then for sure, if you pick $\epsilon > 0$ too large it may happens that $(c-\epsilon , c+\epsilon)$ won't be contained in $A^c$ because it won't be contained in $(1,\infty)$. For instance, take $\epsilon = c+1$. But you can take $\epsilon > 0$ small enough, such that $$ (c-\epsilon , c+\epsilon) \subset (1,\infty). $$ For instance, $\epsilon = (c-1)/2$ works. And this is actually all we want or need!