Does an Open Cover of Rationals in [0,1] Necessarily Encompass All Irrationals?

[Bashyboy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

In the course of solving this problem, I conjectured that if $\{I_k\}_{k=1}^n$ covers $B$, then surely it must cover $[0,1]$. After a few attempts at proving this conjecture, it suddenly it occurred to me that taking the closure of $B$, and using the fact that $m^*(\overline{I_k}) = m^*(I_k)$, would knock of this problem, and so I solved it this way. However, it still leaves me wondering whether my conjecture is true. I could use a hint on how to prove it. Obviously $\bigcup I_{k}$ must contain some irrational numbers, since each $I_k = (a_k,b_k)$ contains infinitely many rationals and irrationals between the endpoints.

 

[Dick]

Homework Statement

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

Homework Equations

The Attempt at a Solution

In the course of solving this problem, I conjectured that if $\{I_k\}_{k=1}^n$ covers $B$, then surely it must cover $[0,1]$. After a few attempts at proving this conjecture, it suddenly it occurred to me that taking the closure of $B$, and using the fact that $m^*(\overline{I_k}) = m^*(I_k)$, would knock of this problem, and so I solved it this way. However, it still leaves me wondering whether my conjecture is true. I could use a hint on how to prove it. Obviously $\bigcup I_{k}$ must contain some irrational numbers, since each $I_k = (a_k,b_k)$ contains infinitely many rationals and irrationals between the endpoints.

Your conjecture is not true. The rationals are countable so let $\{r_k\}$ be an enumeration. Then let $I_k$ be the ball around $r_k$ of radius $1/2^{k+2}$. What's the total length?

 

[fresh_42]

Your conjecture is not true. The rationals are countable so let $\{r_k\}$ be an enumeration. Then let $I_k$ be the ball around $r_k$ of radius $1/2^{k+2}$. What's the total length?

What's the finite subcover?

 

[Dick]

What's the finite subcover?

Why should there be a finite subcover? $B$ isn't compact.

 

[fresh_42]

Why should there be a finite subcover? $B$ isn't compact.

Yes, but there are only $n$ sets $I_k$ allowed to cover $B$.

 

[Dick]

Yes, but there are only $n$ sets $I_k$ allowed to cover $B$.

Right. My reading is sloppy. But still Bashyboy's conjecture isn't true. If it's an open cover, you can certainly leave out some isolated irrationals.

 

[fresh_42]

Right. My reading is sloppy. But still Bashyboy's conjecture isn't true. If it's an open cover, you can certainly leave out some isolated irrationals.

I'm not convinced, that this will be sufficient. What should an isolated irrational be? $B$ is dense in $[0,1]$.
I cannot see a flaw in @Bashyboy's argument, as $[0,1]=\overline{B}\subseteq \overline{\cup I}= \cup\overline{I}$ and we get $1 \leq m^*(\cup \overline{I}) \leq \sum m^*(\overline{I}) = \sum m^*(I)$.

 

[Dick]

I'm not convinced, that this will be sufficient. What should an isolated irrational be? $B$ is dense in $[0,1]$.
I cannot see a flaw in @Bashyboy's argument, as $[0,1]=\overline{B}\subseteq \overline{\cup I}= \cup\overline{I}$ and we get $1 \leq m^*(\cup \overline{I}) \leq \sum m^*(\overline{I}) = \sum m^*(I)$.

There's nothing wrong with that argument. The conjecture was that if the $I_k$ cover the rationals in $[0,1]$, then they cover the whole interval. That's the part that's false.

 

[Ray Vickson]

There's nothing wrong with that argument. The conjecture was that if the $I_k$ cover the rationals in $[0,1]$, then they cover the whole interval. That's the part that's false.

I don't see why.

Let $x \in (0,1)$ be an irrational, and let $\{ q_k \}]$ be a sequence of rationals in $[0,1]$ that converges to $x$. There are intervals $\{ I(q_k)\} \subset B$ that cover the $q_k$, but since the cover $B$ is finite, there is some single $I \in B$ that covers all $q_k$ for $k > K$ ($K$ finite). In other words, we have some interval $I = (a,b) \in B$ with rational endpoints such that $a < q_k < b$ for all $k > K$. Therefore, the limit $x$ lies in $(a,b)$ also. (We have that $x \in [a,b]$, but $x$ cannot equal $a$ or $b$ because these are rational while $x$ is not.)

 

[Dick]

I don't see why.

Let $x \in (0,1)$ be an irrational, and let $\{ q_k \}]$ be a sequence of rationals in $[0,1]$ that converges to $x$. There are intervals $\{ I(q_k)\} \subset B$ that cover the $q_k$, but since the cover $B$ is finite, there is some single $I \in B$ that covers all $q_k$ for $k > K$ ($K$ finite). In other words, we have some interval $I = (a,b) \in B$ with rational endpoints such that $a < q_k < b$ for all $k > K$. Therefore, the limit $x$ lies in $(a,b)$ also. (We have that $x \in [a,b]$, but $x$ cannot equal $a$ or $b$ because these are rational while $x$ is not.)

Doesn't $\{(-1,x), (x,2)\}$ cover all the rationals but not the irrational $x$?

 

[Ray Vickson]

Doesn't $\{(-1,x), (x,2)\}$ cover all the rationals but not the irrational $x$?

Oh, OK: I mis-read the problem specification and made the intervals $I_k$ rational, but that was not part of the original problem statement.