number theory - Primes and arithmetic progressions

In a book on complex analysis, the authors prove:

Given finitely many (non-trivial) arithmetic progressions of natural numbers $$a_1, a_1+d_1, a_1+2d_1, \cdots$$
$$a_2, a_2+d_2, a_2+2d_2, \cdots,$$
$$a_k, a_k+d_k, a_k+2d_k, \cdots,$$
their totality (union) is never $\mathbb{N}$.

Given: any $k$ (non-trivial) arithmetic progressions.

Let $S$ denote the set of those numbers which are not in any of above $k$ arithmetic progressions.

If $S$ would have been finite, then we can form new finitely many arithmetic progressions which covers $S$, and combining them with given progressions, we will cover $\mathbb{N}$ by finitely many arithmetic progressions, a contradiction.

Thus $S$, the complement of all the given $k$ progressions, must be infinite.

Question Does $S$ contain infinitely many primes?

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A non-trivial arithmetic progression means, the common difference is not $1$, i.e. it is not of the type
$$a, a+1, a+2, \cdots$$