Infinite Subsets!

Yes, this is indeed possible. Call the 2 sets $A$ and $B$ . Any positive integer with an odd number of digits (eg 9, 232, 3726762) shall go in $A$ , and any positive integer with an even number of digits (eg 12, 4566, 909945) shall go in $B$ .

These two subsets will have no infinitely long arithmetic progressions (Try to prove this).

Providing a proof for the sets A and B that Sharky mentions.

Note that for any pair of natural numbers $m>0$ and $c$ , the (ordered) set $AP_{m,c}=\{km+c| k \in \N\}$ forms an infinite-length arithmetic progression, and any arithmetic progression of infinite length that is a subset of the natural numbers, can be described as such a set.

• Choose any pair of natural numbers $(m,c), m>0$ . The arithmetic progression $AP_{m,c}$ starts at $0m+c=c$
• Pick an odd power p of 10 that is larger than m, e.g. if m=40, then p=1000
• Let q be the first power of 10 that is larger than c.

If $c$ has an odd number of digits, it is in set A. The number $pc$ will have an even number of digits, and is part of a range of pq consecutive numbers with the same number of digits. Since $pq \ge p \gt m$ , this range is larger than m. Look for a k such that $km+c \le pc \lt (k+1)m+c$ Then at least one of $km+c$ and $(k+1)m+c$ , is in set B. Both numbers belong to the arithmetic progression, so the arithmetic progression has one or more members in both A and B. And since $A \cap B = \varnothing$ , this means that neither A nor B contains all members of $AP_{m,c}$ .

Similarly, if $c$ has an even number of digits, it is in set B and either $km+c$ or $(k+1)m+c$ must be in set A .

So the infinite arithmetic progression associated with the pair (m,c) have members in either set. Since we chose any pair (m, c) no infinite length arithmetic progression is a subset of either set A or B.

Note: There are infinitely many partitions that do the trick. For example, we could do the same thing in binary, or any other base, or we could choose the following partition: A contains all numbers $x$ so that $(n-1)^2 < x <= n^2$ for odd values of n, and B contains all numbers x so that ((n-1)^2 < x <= n^2) for even values of n . These are all examples of partitions which alternate chunks of subsequent numbers between the subsets, having increasingly large chunks as we go up. If this is the case it is quite easy to show that from the point where the chunksize exceeds both c and m, every next chunk contains at least one element of the arithmetic progression.

Make set $A$ which is the set of primes { $0,2,3,5,7,11,13…$ }, which is said to have no consistent pattern, and the other set( $B$ ) is the composite ones is { $4,6,8,9,12,14…$ } which, if primes don't have a consistent pattern, shouldn't have one either.