I want to show that, if $\ A \subset \mathbb{N}\ $ has arithmetic progressions of length $\ k\ $ for all positive integers $\ k,\ $ then so do $\ \left\lceil cA \right\rceil := \{\ \left\lceil ca \right\rceil: a\in A \}\ $ and $\ \left\lfloor cA \right\rfloor := \{\ \left\lfloor ca \right\rfloor: a\in A \}\ $ for any real $\ c>0.$
This should be elementary, but I'm having trouble with the details of a proof. We cannot simply use Szemerédi's theorem because a set with arithmetic progressions of any length need not have positive upper density. So an elementary proof would look something like this:
Let $\ k\in \mathbb{N}.\ $ Then $\ A\ $ has an arithmetic progression of length $\ k;\ $ let us focus on one such arithmetic progression $\ S(k):= \{ b, b+d, \ldots, b+(k-1)d\ \}.\ $ As $\ S\ $ depends on $\ k,\ $ I have written $S$ as a function of $\ k,\ $ which I will come back to later, but for now, $\ k\ $ is the originally chosen integer. Then, $\ \left\lceil cS(k) \right\rceil = \{\ \left\lceil cx \right\rceil: x\in S(k) \} = \{\ \left\lceil cb \right\rceil, \left\lceil c(b+d) \right\rceil, \ldots, \left\lceil c(b+(k-1)d \right\rceil \}. \ $ We can assume WLOG that $\ c<1,\ $ because $\ \left\lceil cS(k) \right\rceil\ $ has the same structure of arithmetic progressions (but with different numbers) as $\ \left\lceil \{c\}S(k)\right\rceil.\ $ [Here, $\ \{\cdot\}\ $ means fractional part]. We need to show that as $\ k\to\infty,\ $ the longest arithmetic progression of $\ \left\lceil cS(k) \right\rceil\ $ tends to infinity also.
This is where I'm stuck and not sure what to do. I have some vague ideas about splitting $\ [0,1)\ $ up into intervals $\ \left[0,\frac{1}{d}\right), \left[\frac{1}{d},\frac{2}{d}\right), \ldots, \left[\frac{d-1}{d},1\right),\ $ and seeing which interval $\ c\ $ is in, and/or using Pigeonhole Principle, but it's not all coming together...
Thanks in advance for your help.