Cool problem! (1 Viewer)

largarithmic · Mar 12, 2012

Since there's been a mood for cool problems recently, here's one. It's pretty hard but neat ^^

Let S be a (possibly infinite) collection of open intervals such that the union of S covers the closed interval [0,1]. Prove that there is a subset T of S with finitely many elements such that the union of T also covers [0,1].

Notation:
An open interval between a and b is the set of reals x such that a < x < b, and is denoted by (a,b).
A closed interval between a and b is the set of reals x such that a <= x <= b, and is denoted [a,b].

seanieg89 · Mar 12, 2012

That assertion follows from the fact that [0,1] is a compact subset of R. This is an immediate consequence of the Heine-Borel Theorem. The below direct argument is essentially a special case of how we can prove the Heine-Borel Theorem for boxes in R^n.

Suppose [0,1] has an open covering A of intervals with no finite subcover.

[0,1]=[0,1/2] U [1/2,1], so at least one the intervals on the RHS (which are both covered by A) must have no finite subcover.

We can now repeat this argument for this interval to obtain an interval on length 1/4 with the same property. Etc.

By induction, this gives us an infinite sequence of nested closed intervals, their lengths successively halving. By Cantors intersection theorem (a pretty straightforward consequence of completeness), there is exactly one point x in the intersection of these nested intervals. This point x must lie in some open interval I in the collection A. But from the definition of x (and of open sets in a metric space!), there must exist a small closed interval from our nested sequence that contains x and is a subset of I, (and hence is covered by I).

This is absurd, as we earlier asserted that every closed set in our nested sequence had no finite subcover!

Hence the covering A of [0,1] has a finite subcover.

Carrotsticks · Mar 12, 2012

I'll take a shot at it, though I admit I know little about set theory (having covered very little of it in Statistics last semester). Here are my definitions.

I am quite convinced that there is a flaw somewhere here. I'm not sure if it can be assumed that there are no 'gaps' within S. How I see if *with included assumption* is that:

$\\a_1 < a_2 < ... < a_n $ and $ b_1 < b_2 < ... < b_n$

Not only that, but also

$\\a_2 < b_1 $ and $ a_3 < b_2 $ and $ ... $ and $ a_n < b_{n-1} $ and $ b_1 < b_2 < ... < b_n$

$\\ $Define the general open interval in S to be $ x_n $ where $ n \in \mathbb{N} $ and $ x_n $ is the open interval $ (a_n,b_n). \\\\ $We choose w.l.o.g $ a_1, a_2,...,a_n $ and likewise for b such that $ a_1<a_2<a_3<...<a_n\\\\ $So it follows that: $\\\\ S = \{(a_1,b_1) \cup (a_2, b_2) \cup ... \cup (a_n, b_n) : n \in \mathbb{N}\} = \{ x_1 \cup x_2 \cup ... \cup x_n : n \in \mathbb{N}\}. \\\\ $Define the closed interval $ [0,1] $ to be $ R $ so $ R = [0,1] \\\\ $As stated by the question, the union S covers the interval [0,1], so $ R \subseteq S \\\\ $Define the minimal set within S covering the interval [0,1] to be M such that $ M = \{x_m,x_{m+1},...,{x_p} \} $ for some m and p such that $ m<p; m,p \in \mathbb{N} $ and $ R \in (a_m, b_p). $ This directly implies that $ a_m < 0 $ and $ b_p > 1.$

$\\$We want to show that there exists a subset of S, namely T, that is equal to M. $$

Gotta start Vector Calc tute now.

And read over Analysis notes.

Cool problem! (1 Viewer)

largarithmic

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seanieg89

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Carrotsticks

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