For any unbounded set of real numbers, is there a subset which almost coincides with a uniformly spread out set of points an infinite amount of times?

I figured out yesterday that, given an unbounded infinite sequence $\ (\alpha_n)_{n\in\mathbb{N}}\subset \mathbb{R}\ $ with $\ \alpha_i \neq \alpha_j\ $ if $\ i\neq j\ $, there does not necessarily exist $\ a,b\in\mathbb{R}\ $ with $\ 0\leq a<b\ $ such that $$(a + nb) = \alpha_k$$

for infinitely many distinct pairs $\ (n,k)\ $ where $\ n\in\mathbb{Z}\ $ and $\ k\in\mathbb{N}.$

For a counter-example is the sequence $\ \{\pi^n: n\in\mathbb{N}\}.\ $ For a proof that this is a counter-example, we suppose by contradiction that such $\ a,b\ $ do exist. Then we have:

$a+n_1 b = \pi^{k_1}\quad (1)$

$a+n_2 b = \pi^{k_2}\quad (2)$

$a+n_3 b = \pi^{k_3}\quad (3)$

But then by evaluating $\ \frac{(3)-(2)}{(2)-(1)}\ $ we must have that $\ \pi\ $ is algebraic and not transcendental, which is false: thus we have have our contradiction, and so no such $\ a,b\ $ exists.

This answers the question for whether or not, for every infinite set of real numbers, there is a subset that coincides with a uniformly spread out set of points: the answer is no, and as I just showed, some sets do not even have three points that coincide with a uniformly spread out set of points.

So to me, the following follow-up question is very natural:

$$$$

Given an unbounded infinite sequence $\ (a_n)_{n\in\mathbb{N}}\subset \mathbb{R}\ $, does there exist an arithmetic sequence $\ a+nb\ $ containing a subsequence arbitrarily close to $\ a_k\ ?$

Or another way to ask the same question:

Given a countably infinite set $\ X\subset \mathbb{R},\ $ do there exist $\ a,b\in\mathbb{R}\ $ such that for any $\ \varepsilon>0,\ \text{dist} \left( \{ a+bn: n\in\mathbb{N}\}, x \right)\ < \varepsilon\ $ for infinitely many $\ x\in X\ ?$

It would be strange if the answer to the question was, "no". On the other hand, I don't know how to prove the answer is "yes". So... I am stuck. But since I have taken a break from this question, I feel like we should be able to use some argument with the pigeonhole principle.

Solution 1:

I just realised that the answer is an easy yes, and my answer makes the question look stupid, which it is. I answer the second question, but the first question is essentially the same (but the second question allows for $X$ to be bounded, in which case the answer is trivial by B-W also...).

In fact the answer holds for all choices of $\ b.$

To demonstrate, let's take $\ b=1.$ Then consider the function $\ f: X\to [0,1)\ $ defined by $\ f(x) = x\mod 1.\ $ Then by Bolzano-Weierstrass Theorem, $\ U = \{f(y): y\in X\}\ $ has a limit point somewhere in $\ [0,1)\ $ because $\ X\ $ is infinite and the range of $\ U\ $ is bounded in $\ [0,1).\ $ Call this limit point $\ s.\ $ Then it is clear that $\ a=s,\ b=1\ $ gives an affirmative answer to the question, i.e. for any $\ \varepsilon>0,\ \text{dist} \left( \{ s+n: n\in\mathbb{N}\}, x \right)\ < \varepsilon\ $ for infinitely many $\ x\in X.$