# Removing Open Balls $B(x_i,\frac{1}{2^i})$ from Real Numbers Centered at Every Rationals x(i) where i is in N

The reason I am asking this because even though by measure or length concept it is true.But as Q is dense in R it seems to me that the remaining set is empty after removing open balls
$$B(x_i,\frac{1}{2^i})$$ from R.  So How should I think so that it does not looks like a empty set. Or what i am thinking wrong.

As it's already been pointed out, density is not sufficient to guarantee that those balls cover the whole space. In fact, $$2^{1/2}\notin\bigcup_{q\in\Bbb Q} B(q,\lvert q-2^{1/2}\rvert)$$, but since I am making it painfully obvious that some correlation between the centers of the balls and their radii prevents $$2^{1/2}$$ from being there, you don't find it all that surprising. But $$\Bbb Q$$ is still the same dense set.

Without invoking tools of measure theory, it is still possible to prove that for any $$a\in\Bbb R\setminus \Bbb Q$$ and for any strictly decreasing sequence $$\varepsilon_n\searrow 0$$ there is an enumeration $$f:\Bbb N\to \Bbb Q$$ of the rationals such that $$\lvert f(n)-a\rvert>\varepsilon_n$$ for all $$n\in\Bbb N$$. The essential point is that the usual bijection \begin{align}\Phi:\Bbb N^2&\to\Bbb N\\ \Phi(x,y)&=\frac{(x+y)(x+y-1)}2+x+1\end{align} satisfies $$\max\{x,y\}\le \Phi(x,y)$$ - or, equivalently, that both the components of $$\Phi^{-1}(n)$$ are smaller or equal to $$n$$. Now, consider $$q:\Bbb N\to\Bbb Q$$ any bijective function and define $$U_n=\{k\in\Bbb N\,:\, \varepsilon_n< \lvert q(k)-a\rvert\le\varepsilon_{n-1}\}$$ (with $$\varepsilon_{-1}=\infty$$ out of convenience). It is apparent that every $$U_n$$ is infinite, and therefore each $$U_n$$ is canonically arranged in an infinite strictly increasing sequence $$(b^n_0, b^n_1, b^n_2,\cdots)$$. Moreover, $$U_n\cap U_m=\emptyset$$ for all $$m\ne n$$, therefore the map \begin{align}\Psi:\Bbb N^2&\to \Bbb N\\ (n,h)&\mapsto b^n_h:=\text{the }h\text{-th smallest element of }U_n\end{align} is bijective.

I claim that $$f=q\circ \Psi\circ \Phi^{-1}$$ is a sound candidate. In fact, if $$(x_n,y_n)=\Phi^{-1}(n)$$, then by definition $$\Psi(\Phi^{-1}(n))=\Psi(x_n,y_n)\in U_{x_n}$$. Also by definition $$\lvert q(\Psi(\Phi^{-1}(n)))-a\rvert>\varepsilon_{x_n}$$. But $$x_n\le n$$, therefore $$\varepsilon_{x_n}\ge \varepsilon_n$$ and thus $$\lvert q(\Psi(\Phi^{-1}(n)))-a\rvert>\varepsilon_n$$

QED.

• Remark: surjectivity of $\Psi$ needs a bit of attention, because that's where you end up using the facts that $\varepsilon_n\to 0$ and that $a\notin \Bbb Q$; it is ultimately true, though.
– user239203
Sep 26, 2019 at 7:14

If you pick a specific real number $$x$$, there is no reason to assume that $$\lvert x-x_i\rvert<\frac1{2^i}$$, for some $$i\in\mathbb N$$. In other words, there is no reason to assume that $$x\in B\left(x_i,\frac1{2^i}\right)$$ for some $$i\in\mathbb N$$. So, the fact that $$\mathbb Q$$ is dense in $$\mathbb R$$ doesn't allow you to deduce that $$\bigcup_{i\in\mathbb N}B\left(x_i,\frac1{2^i}\right)=\mathbb R$$.

• But you are saying for some i that means there are some i ,for that it could follows . I mean can you give some irrationals that are remaining in this set after removing these balls Sep 25, 2019 at 13:26
• No, I cannot, because that depends on the choice of the sequence $(x_i)_{i\in\mathbb N}$. Sep 25, 2019 at 13:30
• If you choose your own sequence Sep 25, 2019 at 13:31
• I will not spend my time defining an enumeration of $\mathbb Q$. If you provide one, I will think about it. Sep 25, 2019 at 13:34
• @Rajat: Problem B1 on the 9th Putnam Exam (March 1949) asks one to prove $\frac{\sqrt{2}}{2}$ is not included in the union of the countably many closed intervals centered at rationals $p/q$ (relatively prime representation), having length $\frac{1}{2q^2}.$ The main idea is to use irrationality of $\sqrt 2$ (via $|q^2 - 2p^2| \geq 1)$ to show that $\left|\left(\frac{\sqrt{2}}{2} - \frac{p}{q}\right)\left(\frac{\sqrt{2}}{2} + \frac{p}{q}\right)\right| \geq \frac{1}{2q^2}.$ This is probably worked out somewhere in Mathematics Stack Exchange, but I couldn't find it. Sep 25, 2019 at 16:22