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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.

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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.

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    $\begingroup$ 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. $\endgroup$
    – user239203
    Sep 26, 2019 at 7:14
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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 $\bigcup_{i\in\mathbb N}B\left(x_i,\frac1{2^i}\right)=\mathbb R$.

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  • $\begingroup$ 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 $\endgroup$
    – Rajat
    Sep 25, 2019 at 13:26
  • $\begingroup$ No, I cannot, because that depends on the choice of the sequence $(x_i)_{i\in\mathbb N}$. $\endgroup$ Sep 25, 2019 at 13:30
  • $\begingroup$ If you choose your own sequence $\endgroup$
    – Rajat
    Sep 25, 2019 at 13:31
  • $\begingroup$ I will not spend my time defining an enumeration of $\mathbb Q$. If you provide one, I will think about it. $\endgroup$ Sep 25, 2019 at 13:34
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    $\begingroup$ @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. $\endgroup$ Sep 25, 2019 at 16:22

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