Prove sum of the lengths of intervals in a finite covering of $\mathbb{Q}\cap [0,1]$ is $\geq 1$

Solution 1:

Let $B$ be the set of rational numbers in the interval $[0,1]$, and let $\{I_k\}_{k=1}^n$ be a finite collection of open intervals that covers $B$. Prove that $\sum_{k=1}^n m^*(I_k) \geq 1$.

Proof:

First, we have $B = \mathbb{Q}\cap [0,1] \subset [0,1]$ and $B \subset \bigcup_{k=1}^n I_k$, where each $I_k$ is an open interval. It follows immediately that since $B$ is countable has outer measure zero and $$B \subset [0,1] \Rightarrow m^*(B) \leq m^*([0,1]) = 1,$$ $$B \subset \bigcup_{k=1}^n I_k \Rightarrow m^*(B) \leq m^*(\bigcup_{k=1}^n I_k) \leq \sum_{k=1}^{\infty} m^*(I_k).$$ Now, since $0 \in B$, there exist one of the $I_k$'s that contains $0$, denote such interval by $(a_1,b_1)$. By a similar reasoning, there exist one of the $I_k$'s that contains $1$ (Note that $1 \in B$), denote such interval by $(a_N,b_N)$. We thus have: $$a_1 < 0 < b_1$$ $$a_N < 1 < b_N$$ Now, lets assume the contrary, that is $\sum_{k=1}^n m^*(I_k) < 1$. In particular, we are assuming the following: $$\sum_{k=1}^n m^*(I_k) < m^*([0,1])$$ We claim that if $b_1 \geq 1$, we obtain a contradiction, since: $$a_1 < 0 < 1 \leq b_1$$ and $$ l((a_1,b_1)) = b_1 - a_1 \leq \sum_{k=1}^n m^*(I_k) < 1,$$ but $b_1 - a_1 > 1$. Otherwise, $b_1 \in [0,1)$ and since $b_1 \notin (a_1,b_1)$, there exist an interval in the collection $\{I_k\}_{k=1}^{n}$ which we label as $(a_2,b_2) \ni b_1$ and $$a_2 < b_1 < b_2.$$ We claim that if $b_2 \geq 1$, we obtain a contradiction, since: $$b_2 - a_1 < (b_2 -a_2)+(b_1 -a_1) \leq \sum_{k=1}^n m^*(I_k) < 1$$ but $b_1 -a_0 > 1$ because $$a_0 < 0 < 1 \leq b_1.$$ We can continue this selection process until it terminates, as it must since there are only $n$ intervals in the collection $\{I_k\}_{k=1}^{n}$. We thus obtain a sub-collection $\{(a_k,b_k)\}_{k=1}^N$ of $\{I_k\}_{k=1}^{n}$ for which $a_1 < 0,$ while $a_{k+1} < b_k$ for $1 \leq k \leq N-1$ and $b_N>1$. We thus have: $$\begin{align*} b_N-a_1 &< b_N - (a_N-b_{N-1})- \cdots-(a_2-b_1) -a_1 \\ &\leq (b_N-a_N)+(b_{N-1}-a_{N-1})+\cdots +(b_1-a_1) \\ &= \sum_{k=1}^N l(I_k) \leq \sum_{k=1}^n m^*(I_k)<1 \end{align*}$$ a contradiction, since $$l((0,1)) = 1 < b_N-a_1$$ and $$a_1 < 0 < 1 < b_N.$$ Thus, we conclude that: $$\sum_{k=1}^n m^*(I_k) \geq 1.$$

Solution 2:

Your proof does not look right because the key point is that there are only finitely many intervals - this is not true for countably many intervals.

In your case, arrange the intervals $I_k = (a_k,b_k)$ in the obvious way such that $0\in (a_1,b_1), 1\in (a_n,b_n)$. Then, since the rationals are dense, $a_{k+1} \leq b_k$ for all $k$. Replacing $I_k$ by smaller intervals if need be, we may assume that $a_{k+1} = b_k$ for all $k$. Then $$ \sum_{i=1}^n l(I_k) $$ becomes a telescoping sum equal to $b_n - a_1 \geq 1$

Solution 3:

All you need is this (which, by the way, is not true for an infinite sequence of intervals):

If $\sum_{k=1}^{n}l(I_{k})< 1$, then there is an interval $J \subset [0,1]$, of strictly positive length, which has no elements in common with any $I_k$

This $J$, with strictly positive length, must contain rational numbers, which means that the $I_k$ don't cover $B$.

Solution 4:

Consider the corresponding closed intervals. Their union covers $\mathbb{Q}\cap[0,1]$ and is closed ( a finite union of closed sets). Therefore their union contains $[0,1]$. So the sum of the length of the intervals is $\ge 1$.

Solution 5:

$A=\mathbb{Q}\cap[0,1] $
This is a sequence of inequalities.
As $A$ is countable so $m^*(A)=0$. Let $\{I_n\}$ be a finite sequence of intervals covering $A$
$1=m^*([0,1])=m^*(\bar{A})\leq m^*(\bar{\cup I_n})\leq m^*(\cup{\bar{I_n}})\leq \sum m^*(\bar{I_n})=\sum l(\bar{I_n})=\sum l(I_n)$
$m^*$ is the outer measure of a set. Hence the result follows.