# Finite subsets of $\mathcal{U}$ don't cover $C \setminus \{p\}$

I'm familiar with this concept and it makes sense, but the proof for it is eluding me.

Let $p \in C$ and consider the set:

$\mathcal{U}=\{\operatorname{ext}(a,b)\mid p\in (a,b)\}$

Therefore no finite subset of $\mathcal{U}$ covers $C \setminus \{p\}$.

It makes sense that a finite number of exteriors will never cover the continuum $C$ ($C$ being nonempty, having no first or last point, ordered ($a<b$), and connected) without $p$ since $p$ will be in exactly none of the subsets $\mathcal{U}' \subset \mathcal{U}$. I'm not sure if that's enough (or maybe even true) though. If anyone can point me in the right direction it will be very much appreciated. (Also, note that $\operatorname{ext}(a,b)$ is the same as $C \setminus [a,b]$. This clarification is simply based on notation.)

-
What do you mean by $\operatorname{ext}(a,b)$? $C\setminus(a,b)$? – Brian M. Scott Nov 7 '12 at 3:20
I mean $C \setminus ((a,b) \cup {a} \cup {b})$. Sorry for not clarifying beforehand. – Casquibaldo Nov 7 '12 at 3:26
Same thing: $a$ and $b$ are elements of $C\setminus(a,b)$ already, since $(a,b)=\{x\in C:a<x<b\}$. – Brian M. Scott Nov 7 '12 at 3:27
Oh, but I meant to exclude them. That's why I put the parenthesis out. I mean to say that the exterior of $(a,b)$ won't include $a$ or $b$. – Casquibaldo Nov 7 '12 at 4:12
Sorry, I misread you previous comment. What you want, in more standard notation, is $C\setminus[a,b]$. – Brian M. Scott Nov 7 '12 at 4:14

$\newcommand{\ms}{\mathscr}\newcommand{\ext}{\operatorname{ext}}$Let $\ms{F}$ be a finite subset of $\ms U$, say $\ms F=\{\ext(a_1,b_1),\dots,\ext(a_n,b_n)\}$. Let $a=\max\{a_1,\dots,a_n\}$ and $b=\min\{b_1,\dots,b_n\}$; by hypothesis $a<p<b$. To show that $\ms F$ does not cover $C$, it suffices to show that $(a,b)\ne\{p\}$, i.e., that at least one of the open intervals $(a,p)$ and $(p,b)$ is non-empty. In fact both are non-empty, and you can show this simply by showing that
$\qquad\qquad\qquad\qquad\qquad$if $x,y\in C$ and $x<y$, then $(x,y)\ne\varnothing$.
This follows from the fact that $C$ is connected, though the details of the argument depend on just what definition of connectedness you’re using.