Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
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

1 Answer 1

$\newcommand{\ms}{\mathscr}\newcommand{\ext}{\operatorname{ext}}$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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.