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.
-
$(-\infty,a)\cup(b,\infty)$ is open, with the usual topology for $\mathbb R$ – J. W. Tanner Nov 19 '23 at 20:12
-
1If $c\in A^c=(-\infty,a)\cup(b,\infty)$ then, for $\epsilon>0$ small enough (smaller than the distance $\min(|c-a|,|c-b|)$ from $c$ to $A^c$), the set $(c-\epsilon,c+\epsilon)$ will be entirely contained in $A^c$ – Anne Bauval Nov 19 '23 at 20:12
-
$\epsilon$ can be chosen arbitrarily small. also, please note that $c$ is strictly smaller than $a$ or strictly greater than $b$ – niobium Nov 19 '23 at 20:13
-
1Does this answer your question? Let $a$ and $b$ be arbitrary real numbers with $a<b$. Show that $[a,b]$ is closed by proving its complement is open. – Anne Bauval Nov 19 '23 at 20:17
-
Yes, we do not search for an epsilon such that the neighbourhood spills over, but one small enough such that the neighbourhood is included inside the open set. e..g $\epsilon=\frac{|a-c|}2$ or $\frac{|b-c|}2$ works depending which subset you consider. – zwim Nov 19 '23 at 20:19
-
1@AnneBauval The linked question in your vote to close is the same question but the answer there doesn't really address the OPs confusion. So I offered the obvious answer. And then seconded your vote to close. – Ethan Bolker Nov 19 '23 at 20:23
2 Answers
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 $[a,b]$.
- 95,224
- 7
- 108
- 199
-
1ah I see my interpretation of the definition was incorrect. It in fact is not for all epsilon but rather for some epsilon, the neighbourhood is contained within the open interval. This makes sense, thank you – aort01 Nov 19 '23 at 20:15
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!
- 2,007