Cfrgtkky

Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $xin mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$?

This problem was inspired by an easier version of the problem where we assume the stronger condition that $displaystylesum_{n=1}^infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too.

Let $Omega$ denote the set of $xin mathbb{R}$ that do not satisfy the given property; we claim that $Omega neq mathbb{R}$ and so an $xin mathbb{R}$ with given property exists. Indeed, $Omega={xinmathbb{R} | exists nin mathbb{N} mathrm{such that} |x-r_n| leq a_n}=displaystylebigcup_{n=1}^infty [r_n-a_n,r_n+a_n]$

whose Lebesgue measure $lambda(Omega)leq displaystylesum_{n=1}^infty lambda([r_n-a_n,r_n+a_n])=displaystylesum_{n=1}^infty 2a_n < infty$, so $Omega neq mathbb{R}$.

Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.

No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).

(At the same time, given a sequence $(a_i)_{iinmathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)

For simplicity, let's look at $[0,infty)$ rather than $mathbb{R}$ (this doesn't make a substantive difference).

The key point is the following picture: we chop $[0,infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[sum_{0<jle i}{1over j}, (sum_{0<jle i}{1over j})+{1over i+1})$. Note that each $B_k$ has "diameter" ${1over k+1}$, and hence if $qin B_k$ then the ball around $q$ with radius ${2over k+1}$ covers $B_k$.

Now fix any enumeration of rationals $E=(r_i)_{iinmathbb{N}}$, and pick a sequence $n_i$ ($iinmathbb{N}$) of naturals such that:

• $n_i<n_{i+1}$, and




• $r_{n_i}in B_i$.

Such a sequence must exist since $B_icapmathbb{Q}$ is always infinite.

Finally, let $A=(a_i)_{iinmathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2over i+1}$ for all $i$. The set $${(r_{n_i}-a_i, r_{n_i}+a_i): iinmathbb{N}}$$ covers $[0,infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.

The answer is no. Consider the following open cover of $Bbb R$:

$$mathcal U={ (H_{2n} , H_{2n+4}) | n in Bbb N_{ge 1}} cup { (-H_{2n+4} , -H_{2n}) | n in Bbb N_{ge 1}} cup { (-3, 3) }$$
where $$H_n = sum_{j=1}^n frac 1j$$
is the $n$-th harmonic number. Note that for all $varepsilon >0$ only finitely many elements $U in mathcal U$ satisfy $lambda (U) > varepsilon$.

Let ${ U_{n} }_{n ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= lambda (U_n)$ converges to $0$.

Let ${ r_n }_{n ge 1}$ be an enumeration of rational numbers such that for all $n$
$$r_{n} in U_n$$
holds. Recall that $a_n$ is rational, and it converges to $0$.

Now, for all $x in Bbb R$ there exists $n ge 1$ such that $x in U_n$: thus $|x-r_n| < a_n$.