# 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.)

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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

$\qquad\qquad\qquad\qquad\qquad$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.