# Closed intervals in open sets

• Oct 17th 2012, 06:01 PM
I-Think
Closed intervals in open sets
Let $|U_\alpha|_{\alpha{\in}I}$ be a collection of open sets such that $[0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$
Prove $\exists$ a finite number of sets $U_{\alpha_{1}},...U_{\alpha_{n}}$ such that
$[0,1]\subset{\bigcup^n_{i=1}}U_{\alpha_{i}}$

Need some help understanding the question first.
Clarification
The statement $[0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$ means $\exists$ an open set containing number $0$, another (or the same set) containing number $1$ and for every $x, 0\leq{x}\leq{1}, \exists U_\alpha$ such that $x\in{U_\alpha}$, so this implies $[0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$?
• Oct 19th 2012, 12:50 PM
Plato
Re: Closed intervals in open sets
Quote:

Originally Posted by I-Think
Let $|U_\alpha|_{\alpha{\in}I}$ be a collection of open sets such that $[0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$
Prove $\exists$ a finite number of sets $U_{\alpha_{1}},...U_{\alpha_{n}}$ such that
$[0,1]\subset{\bigcup^n_{i=1}}U_{\alpha_{i}}$

This is a standard question about compact sets.
The way one does really depend on the tools you have.
We do not know that. I assume you know that any nonempty set bounded set has a least upper bound.

Let $B\subseteq(0,1]$ such $x\in B\text{ if and only if }[0,x]$ is covered by a finite collection of the $U_n$.
Because $0\in U_k$ for some $k$ then $\exists t\in B\cap U_k$. So $B$ is not empty and bounded above by 1.
Let $c=\text{Lub}(B)$. So $c\le 1$.

Suppose $c<1$. Then some collection $\bigcup\limits_{n = 1}^N {U_{k_n } }$ covers $[0,c]$.
Say $c\in U_{k_n}$ but $\exists s\in U_{k_n}\cap (c,1)$.

But that means the same finite collection covers $[0,s]$ which contradicts the maximal nature of $c$
Thus $c=1$ and the finite collection covers $[0,1].$
• Oct 20th 2012, 07:42 AM
johnsomeone
Re: Closed intervals in open sets
There is one little change needed.
The statement says that the cover is in open sets, not open intervals, so $s \in U_{k_n}$ doesn't guarantee that [0, s] is covered by that same finite subcover. That's because s could be in a separate open component of $U_{k_n}$ than the component of $U_{k_n}$ containing c.
This is easy enough to account for. You could declare that this was a theorem about covers by open intervals, and then have the general open cover case as an immediate corollary. You could also say that the cover used in this proof was actually a refinement of the original cover that used the open intervals that comprised the open sets of the original cover - and then easily deduce that there's a finite subcover of the original cover, because you've shown that there's a finite subcover of the open intervals that comprised the open sets of the original cover.