32
$\begingroup$

Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ with radius $\epsilon/2^k$.

The total length of the intervals is a geometric progression that sums up to $\epsilon$.

Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, each real number is in one of the open intervals.

Thus the entire real line is covered by the union of the $I_k$, thus $\mathbb{R}$ is a null set with measure zero.

Clearly there is something wrong in the above proof, however I am not sure where is it?

Thanks for any help.

$\endgroup$
  • 3
    $\begingroup$ Someone has the audacity to reduce the whole real line to $0$ :). +1 for you. $\endgroup$ – Paramanand Singh May 12 '16 at 10:39
  • $\begingroup$ @ParamanandSingh Haha thanks $\endgroup$ – yoyostein May 12 '16 at 11:17
  • 1
    $\begingroup$ enumerate the rationals in $[0,1]$ : $q_1,\ldots,q_n,\ldots$. for each $n$ th rational let $A_n = \{ |x-q_n| < c_n \ \mid x \in \mathbb{R} \}$ with $(c_n)$ some summable sequence : $\sum_n c_n < \infty$. clearly if we choose $c_n = \frac{|\gamma-q_n|}{2^n}$ with $\gamma$ any fixed irrational, then $\gamma$ is not in $\bigcup_n A_n$. this is a start : we proved that your claim is false, and that not always "every irrational is in $\bigcup_n A_n$" under your assumptions. $\endgroup$ – reuns May 28 '16 at 7:44
  • 3
    $\begingroup$ Yes, given any real number $x$, I can find arbitrarily close rationals. But maybe as I pick closer and closer rationals, the associated intervals get smaller and smaller, so that I'm never able to find one that covers $x$. $\endgroup$ – Jack M May 28 '16 at 8:20
  • 1
    $\begingroup$ @JackM : that's what I tried to show with my example of how choosing $(c_n)$. but seing intuitively the shape of the complementary set $[0,1] \ \setminus \ \bigcup_n A_n$ seems much more difficult (than proving that in some cases it is not empty) $\endgroup$ – reuns May 28 '16 at 8:47
18
$\begingroup$

Here is a slight variation on your construction.

Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ which does not contain $\pi.$

Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, each real number is in one of the open intervals. In particular, $\pi$ is in one of the open intervals.

Now, how can $\pi$ be in one of the open intervals, when each interval was specifically chosen to exclude $\pi$?

Yes, there are rational numbers arbitrarily close to $\pi.$ The catch is that, as the rationals get closer and closer to $\pi,$ the corresponding intervals get shorter and shorter.

$\endgroup$
  • 1
    $\begingroup$ I still don't get it completely.. Let $x \in \mathbb R$; There exists a rational number $q_k$ such that $|x - q_k| < \varepsilon_k$. So the problem is that $\varepsilon_k$ is in general not always smaller than $\frac \varepsilon{2^k}$? I'm still not sure... $\endgroup$ – Ant May 11 '16 at 8:52
  • $\begingroup$ @Ant Start with $\epsilon_k.$ Since $\epsilon_k\gt0,$ there is a rational $q$ such that $|\pi-q|\lt\epsilon_k.$ Since you've enumerated all the rationals, $q=q_j$ for some $j.$ So you've got $|\pi-q_j|\lt\epsilon_k.$ So what? $\endgroup$ – bof May 11 '16 at 10:17
  • $\begingroup$ Your example demonstrates that this is not possible assuming we specify that each interval does not contain $\pi$. However, assuming we don't make such a specification, as in OP's example, why is it so obvious? Moreover, how do we know that the corresponding intervals get shorter and shorter not only for the odd real number (ex. $\pi$) but, in fact, almost all real numbers. $\endgroup$ – MathematicsStudent1122 May 28 '16 at 11:38
  • $\begingroup$ Thanks, this is much more insightful than most of the other hooing and humming around this neck of the woods... $\endgroup$ – goblin May 1 '17 at 5:20
  • $\begingroup$ @MathematicsStudent1122 I think a full answer to the question would show that any such covering of the rationals would miss most of the reals. Unfortunately, I think the only way to prove that would be "Suppose there is a covering which captures most of the reals; then $\mathbb{R}$ would have measure zero. But this is not true, so contradiction. $\endgroup$ – Ovi Feb 7 '18 at 22:31
14
$\begingroup$

This part

Each real number is arbitrarily close to a rational number since $\Bbb Q$ is dense in $\Bbb R$. Thus, each real number is in one of the open intervals.

is not true. For an arbitrary $x\in\Bbb R$, $x$ doesn't necessary lies in any of $I_k$'s. If you are not convinced, you can try showing that there exists an $I_k$ such that $x\in I_k$, it wouldn't be as easy as you thought.

$\endgroup$
  • $\begingroup$ Can you elaborate more on why this isn't true? $\endgroup$ – Ant May 11 '16 at 7:44
  • $\begingroup$ @Ant OK, I'll add that in my answer. $\endgroup$ – BigbearZzz May 11 '16 at 7:45
9
$\begingroup$

Heine-Borel theorem prevents such covering.

Actually the answer is ridiculously easy. Suppose you can cover irrational and rational numbers contained in an interval $[0, 1]$ with a countable union of small intervals and sum of these intervals is less than $1$ then according to Heine-Borel theorem you can take a finite number of these intervals and should cover $[0, 1]$ so you get a contradiction as the sum of your intervals is less than $1$.

$\endgroup$
  • 5
    $\begingroup$ This shows that the Heine Borel Theorem is not a naive/intuitive theorem. The fact that it leads to finite coverings is something very very valuable because dealing with a finite number of things is so much easier than dealing with infinite number of things. +1 $\endgroup$ – Paramanand Singh May 12 '16 at 10:30
9
$\begingroup$

This is not exactly an answer to your question, but is rather a historical digression and is too long for a comment.

The exact same question was asked by Axel Harnack in 1885 and he used interval $[0, 1]$ in place of whole of $\mathbb{R}$. Harnack convinced himself that his reasoning is correct and the interval $[0, 1]$ can be covered by a countable number of intervals of any desired length $\epsilon$. He believed that if $A$ is a union of a countable number of intervals then complement of $A$ (denoted by $A'$) is also a union of countable number of intervals. BTW the belief is correct if we replace "countable" by "finite".

His idea was that if $A = \bigcup_{k = 1}^{\infty}I_{k}$ then the complement $A' = [0, 1] - A$ is also a countable union of intervals. Moreover each such interval which is part of $A'$ must consist of only a single point, otherwise such intervals will contain an infinity of rational numbers. Thus $A'$ consists of a countable number of points and it can be covered by another set of intervals with total length less than $\epsilon$. Both $A$ and $A'$ together cover $[0, 1]$ and total length of intervals in both $A$ and $A'$ is less than $2\epsilon$.

Nobody thought that Harnack's proof above was flawed till Emile Borel appeared on scene. The assumption that complement of a countable union of intervals is again a countable union of intervals is wrong. Emile Borel proved the following theorem:

If $\{I_{k}\}$ is countable sequence of intervals such that each $I_{k} \subseteq [0, 1]$ and if the total length of intervals $I_{k}$ is less than $1$ then the complement $A' = [0, 1] - \bigcup_{k = 1}^{\infty}I_{k}$ necessarily contains an uncountable number of points.

Borel further mentioned that the above result actually leads to a much stronger result about a countable collection of open intervals namely:

Theorem: If $\{I_{k}\}$ is a countable collection of open intervals whose union contains $[a, b]$ then there is a finite number of these intervals $I_{k}$ whose union contains $[a, b]$.

This marks the birth of Heine Borel Theorem during 1895.

Note: The historical remarks above are taken from an excellent book A Radical Approach to Lebesgue's Theory of Integration by David M. Bressoud which describes the problems mathematicians faced with the concept of real number line and how these problems were finally resolved by Lebesgue using this theory of measure and integration.

$\endgroup$
  • $\begingroup$ @Did: Thanks I realize my mistake :D that indeed helps $\endgroup$ – crskhr May 28 '16 at 7:53
  • $\begingroup$ do you have any idea for understanding the set of irrationals which are not in $\bigcup_k I_k$ ? (with $I_k = \{x \in [0,1] \ \mid |x - q_k| < c_k \}$ for some sequence $(c_k)$ such that $\sum_k c_k < \infty$, and $(q_k)$ an enumeration of the rationals in $[0,1]$) $\endgroup$ – reuns May 28 '16 at 9:12
  • $\begingroup$ @user1952009: I think there is no specific characterization for the irrationals which lie outside the union of $I_{k}$. $\endgroup$ – Paramanand Singh May 28 '16 at 9:24
  • $\begingroup$ @ParamanandSingh : with $c_k(\epsilon) = \epsilon c_k$, it should be possible to understand how the complementary set evolves as $\epsilon \to 0$ ? $\endgroup$ – reuns May 28 '16 at 9:26
  • $\begingroup$ Interesting, nice one! $\endgroup$ – ibnAbu Jun 8 '16 at 0:55
7
$\begingroup$

"Each real number is arbitrarily close to a rational number since ℚ is dense in ℝ . Thus, each real number is in one of the open intervals." Wrong. I'll construct an enumeration specifically to show that your cover does not contain every real number necessarily. $\mathbb{Q}$ is the union of $A_n=\{x\in \mathbb{Q}: \epsilon2^{-n}<|x-\sqrt2|<\epsilon2^{-n+1}\}$ Now, enumerate each $A_n$ as $a_{n,1},a_{n,2},\ldots$. Then, enumerate each $\mathbb{Q}$ as $e_k=a_{m,n-m}$ regularly using each integer $k$ can be uniquely decomposed as $n(n+1)/2 +m$, where $0\le m \le n$. You can see that $\sqrt2$ is not in any of the interval you've constructed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.