4
$\begingroup$

I am trying to solve the problem that asks the following.

For a given sequence $(a_j)$ of positive real numbers, show that there exists an enumeration of rationals $\{r_1,r_2,\ldots\}$ such that $\bigcup_{j=1}^\infty (r_j-a_j,r_j+a_j)=\mathbb{R}$ if and only if $\sum_{j=1}^{\infty}a_j=\infty$.

I showed with ease that if $\sum_{j=1}^{\infty}a_j<\infty$, it cannot cover $\mathbb{R}$. But I am stuck with the converse: I know that there are enumerations such that even when $\sum_{j}a_j=\infty$, the union does not cover $\mathbb{R}$. But how do I show that there always exists one?

$\endgroup$
1
$\begingroup$

Let $0<\epsilon_j<\min\{2^{-j},a_j/2\}$ be such that $0< b_j=a_j-\epsilon_j\in\mathbb{Q}$.

We have that $\sum b_j=\infty$ iff $\sum a_j=\infty$ because $\sum\varepsilon_j\le 1$

You can cover $\mathbb{R}$ with closed intervals $I_j$ of lenght $2b_j$ such that all their endpoints are at rational numbers: $I_1=(0,2b_1)$, $I_2=(-2b_2,0)$, $I_3=(2b_1,2b_1+2b_3)$ etcetera (for some sequences $b_j$ this does not work but you can always find a way)

Choose your $r_j$ to be the middle point of the interval $I_j$ (it is a ratonal number)

Then $I_j=[r_j-b_j,r_j+b_j]\subseteq(r_j-a_j,r_j+a_j)$ so that $$\mathbb{R}=\bigcup\,[r_j-b_j,r_j+b_j]\subseteq\bigcup\,(r_j-a_j,r_j+a_j)$$ $\sum a_j=\infty\implies\exists\{r_1,r_2,\dots\}=\mathbb{Q}$ such that $\mathbb{R}=\bigcup\,(r_j-a_j,r_j+a_j)$

Edit for a shorter proof: Cover the real line with closed intervals $I_j$ of lenght $a_j$ (this is possible because $\sum a_j=\infty$) then choose $r_j$ to be any rational number in the interior of $I_j$, then $I_j\subseteq(r_j-a_j,r_j+a_j)$ and the result follows

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ For what sequence $b_j$ does your construction fail? $\endgroup$ – user134070 Jan 4 at 3:03
  • $\begingroup$ $\{1,1,\frac{1}{2},\frac{1}{4},\frac{1}{3},\frac{1}{9},\dots,\frac{1}{n},\frac{1}{n^2},\dots\}$ if you place the intervals alternatively to the left and to the right of $0$ then in one side you can only cover the lenght $\sum 2/n^2<\infty$ and yet the sum of the whole sequence is $\infty$ $\endgroup$ – augustoperez Jan 4 at 3:06
  • $\begingroup$ So how do you remedy such issues then? You said there is always a way, but is it something easily provable? $\endgroup$ – user134070 Jan 4 at 3:08
  • $\begingroup$ for any sequence you start convering to the right and use as many intervals as needed to cover (0,1). You'll need finitely many intervals. Then do the same to the right to cover (-1,0), again this is possible and can be made with finitely many inervals. Repeat to cover $(1,2)$ then $(-2,-1)$ then (2,3)... $\endgroup$ – augustoperez Jan 4 at 3:10
  • 2
    $\begingroup$ So in a way it amounts to finding subsequences $n_{j_k}$ and $n_{l_k}$ such that both sums diverge, I suppose? Thanks for your help, btw! $\endgroup$ – user134070 Jan 4 at 3:13

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.