Let $\mathbb{P}$ be the set of prime numbers.
My original question was going to be this:
If $A\subset \mathbb{N}$ with $\ \vert A \vert = \infty,\ $ does $\ \exists\ p\in\mathbb{P} $ such that $\ \displaystyle \{ a \pmod p: a\in A \} = [p]:= \{0,1,\ldots,p-1\}\ ?$
But then I realised that $ A= \{ 2^k: k\in\mathbb{N} \}$ is a counter-example, because no such $p\in\mathbb{P}$ exists with the desired property. This is because for $p=2,$ no member of $A$ is $\equiv 1 \pmod 2.$ And for $p\geq 3,\ $ no member of $A$ is $\equiv 0 \pmod p$ because members of $A$ do not contain multiples of primes other than $2.\ $
With similar reasoning, we see that for any prime $p,\ A= \{ p^k: k\in\mathbb{N} \}$ is a counter-example.
The following question naturally arises:
If $A\subset \mathbb{N}$ with $\ \vert A \vert = \infty.\ $ Does $\exists\ p\in\mathbb{P}_{\geq 3}\ $ such that $\ \displaystyle \vert\ \{ a \pmod p: a\in A \}\ \vert = p\ $ or $\ p-1\ ?$
Or much stronger:
For every $M\in\mathbb{N},\ $ does $\exists\ p\in\mathbb{P}_{\geq M}\ $ such that $\ \displaystyle \vert\ \{ a \pmod p: a\in A \}\ \vert = p\ $ or $\ p-1\ ?$
Elementary number theory theorems like Chinese Remainder Theorem come to mind, as does the pigeonhole principle, but I just don't see how to use these yet to answer the new question. Maybe there is a counter-example, but I do not currently see how to construct it.
Let $p_1,p_2,\ldots$ be the primes in ascending order, starting with $p_1=3$ (so $p_2=5$, $p_3=7$, etc).
Let $$\begin{align*} a_0 &=1\\ a_1 &= 1+p_1\\ a_2 &= 1+p_1p_2\\ &\vdots\\ a_k &= 1+p_1p_2\cdots p_k\\ &\vdots \end{align*}$$ and let $A=\{a_0,a_1,\ldots\}$. Let $p\geq 3$ be a prime, say $p=p_n$. For all $m\geq n$ we have $a_m = 1+p_1\cdots p_m \equiv 1\pmod{p_n}$. Thus, the set $\{a\bmod p_n\mid a\in A\}$ has at most $n$ distinct elements (namely, $a_0\bmod p_n,\ldots,a_{n-1}\bmod p_n$). Because $p_n\geq n+2$ holds for all $n$, it follows $p_n-1\geq n+1 \gt n$, so that $|\{a\bmod p_n\mid a\in A\}|\leq n \lt p_n-1$.