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Let $r_k$ be the rational numbers in $\mathbb{R}$.

(1).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2}) = \mathbb{R}$?

(2).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?

(1).Because $m(\mathbb{R})=+\infty, \sum_{k=1}^\infty \frac{1}{k^2}<+\infty$, so $\mathbb{R} \setminus\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2})\neq \Phi $ (2) What about (2)?

Shine
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  • For each k, does $r_k$ is the same rational number? (I guess that no, else its trivial). So for 2 im sure that there's an order such that it's equal to $\mathbb R$ but im also pretty sure that not every order is good enough.. – Snufsan Jul 06 '14 at 11:52

3 Answers3

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One can choose an enumeration $\{r_k\}_{k=1}^\infty$ of the rationals so that (2) fails.

Define $A = \{2^k : k \in \Bbb N\}$.

Let $\{r_k\}_{k=1}^\infty$ be an enumeration of the rationals such that $$ \{k \in \Bbb N : r_k \in \Bbb Q \cap [0, \infty)\} = A. $$

This is possible since $A$, $\Bbb N - A$, $\Bbb Q \cap [0, \infty)$ and $\Bbb Q \cap (-\infty, 0)$ are all countably-infinite.

Note that $[2, \infty)$ cannot be covered by $\bigcup_{r=1}^\infty (r_k - 1/k, r_k + 1/k)$. This is because only the intervals $(r_k - 1/k, r_k + 1/k)$ for $k \in A$ can cover elements in $[2, \infty)$, and the sum of the lengths of those intervals is finite.

Ayman Hourieh
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  • In fact this construction shows that there is an enumeration such that the measure of the union is abritrarily small. Set aside some interval $I$ of length $\epsilon$ and enumerate all rationals outside of $I$ in indices coresponding to a series with sum less that $\epsilon$, then fill the remaining positions with rationals from $I$. – Rene Schipperus Jul 06 '14 at 12:12
  • @ReneSchipperus Nice observation! – Ayman Hourieh Jul 06 '14 at 12:13
  • Where "arbitrarily small" means $> 2$, but as close to $2$ as one wishes. – Daniel Fischer Jul 06 '14 at 12:13
  • @Ayman Hourieh. Yes, in this way of establishment, the union can't cover the whole real line. But if we add the condition that $\mathbb{N}\rightarrow \mathbb{R}: k\rightarrow r_k$ is a bijection. Then what happens? the different enumeration also leads to different result? – Shine Jul 06 '14 at 12:55
  • @Rene: since the first term in the sum has measure 2. – GEdgar Jul 06 '14 at 12:56
  • @GEdgar yes right, I guess didnt understand what was being asked. – Rene Schipperus Jul 06 '14 at 15:08
  • @Shine The condition that $\Bbb N \to \Bbb Q$ is a bijection (note the codomain) is already satisfied by my answer. This is the definition of an enumeration of $\Bbb Q$. Different enumerations lead to different results indeed. I provide an example in which (2) fails. See Rene's answer for an example in which (2) is true. – Ayman Hourieh Jul 06 '14 at 16:17
  • @Ayman Houieh, yes right ! – Shine Jul 07 '14 at 03:47
2

I think (2) depends on how you enumerate the rationals. For example lets say you dont want $e$ in the image. Then enumerate the rationals so that if $q$ is a rational and $e-q \sim \frac{1}{n}$ the make sure that if $r_k=q$ we have $k > n$. (better is given in Ayman's answer and in the comments afterwards). Conversely, there is an enumeration so that (2) is true, for example to cover the unit interval chose the rational sequence increasing so that you succsssively cover the interval, since the series diverges you will cover the whole interval, then do this over every interval.

I have given a very rough description of these constructions, they involve much back and forth to ensure one has a enumeration of all the rationls.

Rene Schipperus
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For $k=2n+1(n \in N)$ we put $q_{2k}=\sum_{i=1}^{n-1}1/(2(2i+1))+1/(4(2n+1))$; For $k=2n(n \in N)$ we put $q_{2k}=-\sum_{i=1}^{n-1}1/(4i)-1/(8n)$;

Let $f:N\setminus \{ 2k:k \in N\} \to Q \setminus \{q_{2k}:k\in N\}$ be a bijection. We put $q_n=f(n)$ for $n \in N\setminus \{ 2k:k \in N\}$. Then for the sequence $(q_n)_{n \in N}$ we have $R=\cup_{n \in N}(q_n-1/n,q_n+1/n)$.

George
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