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

I have the following proof, but I don't think it's right. Could somebody please tell me what's wrong with it, and how to fix it? Thanks :)

Let $$B$$ be the set of rational numbers in the interval $$[0,1]$$ Let $$I_{k}$$, $$k=1,2,\cdots, n$$ be a finite collection of open intervals that covers $$B$$. Prove that $$\sum_{k=1}^{n}l(I_{k})\geq 1$$.

($$l(I)$$ denotes the length of the interval $$I$$).

$$\underline{\mathbf{\text{Proof}}}$$: Suppose $$a,b \in [0,1]$$ such that $$a < b$$. By the density of the rational numbers in $$\mathbb{R}$$, $$\exists q_{i} \in \mathbb{Q}$$ such that $$\forall a_{i} < b_{i}$$, $$q_{i}\in(a_{i},b_{i})$$.

Since the rationals are dense in $$\mathbb{R}$$ and hence in $$[0,1]$$, we can cover $$[0,1]$$ as follows:

$$[0,1] \subseteq \cup_{k=1}^{\infty}I_{k}=\cup_{k=1}^{\infty} (a_{k},b_{k})$$.

(In the case where the $$(a_{k},b_{k})$$ do not completely cover $$[0,1]$$ completely, $$\exists$$ a point between $$(a_{i},b_{i})$$ and $$(a_{i+1}, b_{i+1})$$ that remains uncovered. However, because there are countably many intervals, there will be only countably many such points. Thus, the set of points, call it $$S$$, being a countable set will have $$m^{*}(S)=0$$. So, these points will be negligible in terms of their effect on the sum of the lengths of the intervals.)

By Heine-Borel, since $$[0,1]$$ is closed and bounded, every such open cover has a finite subcover. So, $$[0,1] \subset \cup_{k=1}^{n}(a_{k},b_{k}) = \cup_{k=1}^{n}I_{k}$$.

Now, the outer measure of an interval is its length. So $$m^{*}([0,1])=l([0,1]) = 1$$, and by the definition of the outer measure, $$1 = m^{*}([0,1]) = \inf_{k=1}^{n}l(I_{k}) \leq \sum_{k=1}^{n}l(I_{k})$$

Q.E.D.

I also realize this question has been asked before on here, but I am terribly dissatisfied with the answers that have thus far been given.

• What you have shown is that a cover of [0,1] by open intervals has a finite subcover $F$ such that $m^*(\cup F)\geq 1$ which does not address the question. Commented Oct 13, 2015 at 3:54

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: $$\sum_{k=1}^n m^*(I_k) \geq 1.$$

• I'm a little confused with some of your notation. For example, where do $b_{0}$ and $a_{0}$ come from? Our indices start at $k=1$. Also, when you say "We claim that if $b \geq 1$, we obtain a contradiction" (the very first time you say this), did you mean $b_{1}$?
– user100463
Commented Oct 13, 2015 at 5:00
• yeap i mean $b_1$. I think that insted of $a_0$ and $b_0$ it should be $a_1$ and $b_1$. I need to verify, it should be a typo.
– okie
Commented Oct 13, 2015 at 5:11
• This is fine, but was there any motivation to present the construction with that contradiction? Noting that $1 = m^{*}[0, 1]$ contained in the $n$-th stage of construction if $b_{n} \geq 1$ would alleviate the need for a contradiction at all. Bit more clear for me at least. Commented Jun 5, 2017 at 20:07
• How did you know $b_1$ is rational? Commented Oct 2, 2022 at 19:39

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 $$\ge 1$$.

• would you please explain why "union covers Q∩[0,1] and is closed" can give us "their union contains [0,1]". Thanks!
Commented Sep 2, 2022 at 5:49
• @FaDA: $\mathbb{Q}\cap [0,1]$ is dense in $[0,1]$. It's something about closed sets, dense subsets, closure etc. It's fairly basic, but might not be clear without some topology refresher. Commented Sep 2, 2022 at 5:56
• I see, i will check these concepts for better understanding, thanks!
Commented Sep 2, 2022 at 6:02

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 $b_n - a_1 \geq 1$

• Maybe you have the inequality reversed in the intervals? Because, wouldn't that leave a gap otherwise? Commented Oct 13, 2015 at 2:33
• @Esoog: Thanks, fixed. Commented Oct 13, 2015 at 2:34

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 $B$.

$A=\mathbb{Q}\cap[0,1]$
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.

It's obvious we need only consider bounded open intervals . For any non-empty finite set $C$ of non-empty bounded open intervals let $$d(C)=\min \{a |\exists b [(a,b\in C]\}.$$ $$\text {And then let } e(C)=\max \{b|(d(C),b)\in C\}.$$ $$\text {And for any } D\subseteq C \text { let } D^*=\{s\in C | s \not \subseteq \cup D\}.$$ $$\text {Now let } D_0=\phi \text { and let } D_{n+1}=D_n\cup \{(d(D_n^*),e(D_n^*)\}\text { if } \cup D_n\ne \cup C,$$ $$\text { or } D_{n+1}=D_n \text { if } \cup D_n=\cup C.$$ I leave it to you to show that $$(1)... \exists n (\cup D_n=\cup C)$$ and that if $m$ is the least $n$ satisfying (1) and if $Q\cap [0,1]\subset \cup C$ , then $$(2)...D_{m}=\{(u_j,v_j) : j=1,...,m\}$$ $$\text {where } u_{j+1}\leq v_j<v_{j+1} \text { whenever }1\leq j\leq m-1$$ $$\text {and (of course) } u_1<0\wedge v_{m}>1.$$ $$\text { Since } D_{m}\subset C \text { we have } \sum_{s \in C} l(s)\geq \sum_{s\in D_m} l(s)=\sum_{j=1}^{j=m} (v_j-u_j)\geq$$ $$(v_1-u_1)+\sum_{j=1}^{j=m-1} (v_{j+1}-v_j)=v_m-u_1>1.$$.....Footnote:In (2) we have $(u_1,v_1)=(d(C),e(C))$ , and when $j<m$ we have $(u_{j+1},v_{j+1})=(d(D_j^*,e(D_j^*))$.And in the last "summation",if $m=1$ we take the value of the summation to be $0$.