# union of a collection of open interval containing the rationals in $[0,1]$

I have a question regarding union of some finite collection of open interval (say $\bigcup _{k=1}^n I_i$, where each of $I_i$ is an open interval ) containing the rational numbers in the interval $[0,1]$.

Basically I am trying to show that the summation of each outer measure (i.e. $\displaystyle\sum_{i=1}^{k} m^*(I_k) \geq 1$ ) of the union of the collection of those open intervals has a measure of at least one.

I basically try to use the fact that the outer measure of the irrational (denoted say by $IQ$) has an outer measure of 1 i.e. $m^*(IQ)=1$ in the interval $[0,1]$.

I am guessing that the set of IQ is a subset of $\bigcup _{k=1}^n I_i$, so that by monotonicity and countable sub-additivity of the outer measure, I can say $\sum_{i=1}^{k} m^*(I_k) \geq 1$.

But I have not able to come up with a proof that shows the set of irrational in [0,1] is a subset of the union of the above collection of open intervals yet.

I was trying to show the above subset "guess", by doing this:

that given an irrational number say $iq_i$ in $[0,1]$, then for an arbitrary $\epsilon >0$, we have a rational (say denoted as q) in the interval of $(iq_i -\epsilon, iq_i +\epsilon)$, and then try to say that open interval belongs to one of the $I_k$, but somehow not successful yet as I cannot "concretely" prove it yet (i.e. using delta epsilon style proof). Or other concrete way of doing it yet.

If anyone can give some suggestions, it would be great.

thank you

The thing you're trying to prove is incorrect - e.g. $(-1, {\pi\over 4})\cup({\pi\over 4}, 2)$ is a pair of open intervals which cover $[0, 1]\cap\mathbb{Q}$, but which don't contain every irrational (in particular, they don't contain ${\pi\over 4}$). Your argument is going to have to be more complicated.

In particular, note that if we allow infinite collections of intervals, then we can make do with total measure much less than $1$: cover $q_n$ with an interval of width ${1\over 2^n}$, for some fixed listing $\{q_n: n\in\mathbb{N}\}$ of the rationals. So your argument is going to need to use something special about finite covers.

There are two approaches I know of here:

• Use induction on the number of intervals. This is a bit trickier than it first appears - this is one of those problems where you have to strengthen the inductive hypothesis . . .

Or you can

• Show that every irrational which is not an endpoint of one of the intervals is in the union of the intervals; since there are only finitely many intervals, this means all but finitely many irrrationals are in the union of the intervals, and this is enough to conclude that the total size of the intervals is $\ge 1$.
• LOL haha, such an obvious counterexample to disprove my stupid guess. Thanks for giving the example so quickly. Jan 10, 2017 at 15:33
• @bof . . . Wow. That's not a good sign. Fixed! Jan 11, 2017 at 2:56
• Hi Noah, there seem to be an answer here: math.stackexchange.com/questions/1477462/… but based on what I have read from the chosen answer, do you think there is something wrong in the chosen answer? because the person keep assuming the end point of the interval say $b_j$ of ($a_j, b_j$) would somehow fall into some interval say ($a_{j+1} ,b_{j+1}$) in the proof, but this might not be true, is it correct? Jan 16, 2017 at 7:11

If you have a finite collection of open intervals whose union covers $\mathbb{Q}\cap[0,1]$ then either it covers the entire interval or leaves out at most a finite collection of irrational numbers---all of the interval but for a set of measure $0$.

So the measure of the union of the open intervals cannot be smaller than the measure of the interval $[0,1]$.

• Hi Dr. Bales, could you explain what you means by " leaves out at most a finite collection of irrational numbers". Why only finite of irrational numbers are left out? Jan 10, 2017 at 9:09
• sorry for my stupidity, my ability is not on the same playing ground here yet Jan 10, 2017 at 9:10
• The collection does not have to contain every irrational number in $[0,1]$, it only has to contain every rational in $[0,1]$. Suppose there are only two intervals $I_1,I_2$ such as @Noah Schweber suggested. Then it covers all but one number in $[0,1]$--all but a set of measure $0$. Or what it it has only three intervals $\left(-1,\frac{\pi}{4}\right),\,\left(\frac{\pi}{4},\frac{\pi}{3}\right),\,\left(\frac{\pi}{3},2\right)$? Then it covers all of $[0,1]$ but for $2$ numbers--a set of measure $0$. This can be continued using non-overlapping intervals with endpoints $\frac{\pi}{k}$, $k>4$. Jan 10, 2017 at 17:00

One way to show that, if $F$ is a finite set of open intervals with $\cup F\supset \mathbb Q\cap [0,1],$ then $[0,1]$ \ $\cup F$ is finite:

Let $L=\min \{m(f\cap [0,1]):f\in F \land f\cap [0,1]\ne \phi\}.$ We have $L>0.$

Suppose by contradiction that $S$ is an infinite subset of $[0,1]$ and $S\cap (\cup F)=\phi.$ There exist $s_1,s_2\in S$ with $0<|s_1-s_2|<L.$ There exists a rational $q$ between $s_1$ and $s_2,$ and for some $f\in F$ we have $q\in f.$

But there is an $f\in F$ and an interval $J$ of length $m(J)\geq L,$ such that $q\in J\subset f\cap [0,1].$ Then at least one of $s_1,s_2$ must belong to $J,$ and hence to $f,$ contrary to $\{s_1,s_2\}\cap f\subset S\cap (\cup F)=\phi.$

Notes:

(1). The set $S$ has distinct points as close together as desired, for if $n\in \mathbb N$ and $1/n<L$ then at least one of $\{S\cap [j/n,(j+1)/n]:j=0,...,n-1\}$ has at least $2$ members, otherwise $S$ would be finite.(Pigeon-Hole Principle: If at least $n+1$ pigeons (e.g. members of $S$) fly into $n$ holes, at least one hole has at least $2$ pigeons.)

(2). Why at least one of $s_1,s_2$ must belong to the interval $J$ : Let $a= \min(s_1,s_2)$ and $b=\max (s_1,s_2).$ We have $a<q<b$ and $q\in J$. If $\{a,b\}\cap J=\phi$ then $a\leq \inf J$ and $b\geq \sup J.$ But then $L\leq m(J)=\sup J - \inf J\leq b-c=|s_1-s_2|<L,$ a paradox.

• Hi user, could you explain little bit more about what you mean with the L=min{m(f^[0,1]): f ... Not equal empty set} definition. Is it the length of the shortest open interval in the set F that intersects with [0,1]? Because somewhere in the paragraph below it, you said "But there is an f... and an interval of ..." . Is the f in this second sentence referring to the f in the definition of L? Thank you . Jan 11, 2017 at 1:48
• There may be members of $F$ that don't intersect $[0,1]$ . They're of no interest. When $f\in F$ and $f\cap [0,1]\ne \phi$ then $f\cap [0,1$] is an interval of positive length. And $L$ is the length of the shortest of these...... For ANY $f\in F$ and any $q \in \mathbb Q\cap [0,1]\cap f,$ we have $q\in K=f\cap [0,1]$ where $K$ is an interval of length at least $L,$ so we there's an interval $J$ of length $L$ with $q\in J\subset K.$... I should have said that $q\in J=f\cap [0,1]$ where $J$ is an interval of length $\geq L$ . That would be clearer . So I'm going to change it.Thanks. Jan 11, 2017 at 2:40
• Hi user, I have found an answer it seems, but it might be incorrect. math.stackexchange.com/questions/1477462/… but based on what I have read from the chosen answer, do you think there is something wrong in the chosen answer? because the person keep assuming the end point of the interval say $b_j$ of ($a_j, b_j$) would somehow fall into some interval say ($a_{j+1} ,b_{j+1}$) in the proof, but this might not be true, is it correct? Jan 16, 2017 at 7:12