# A question regarding arithmetic progressions

1. Mar 12, 2012

### drag12

Fairly recently someone started a topic here regarding the conjecture of Erdos about arithmetic progressions, namely that if $A$ is a subset of the natural numbers and the sum of the reciprocals of elements of $A$ diverges, then $A$ contains arbitrarily long arithmetic progressions.

I'm looking for some clarity on the statement mainly, that I haven't been able to find anywhere else. I'll illustrate my question with an example:

Consider the set $A=\{1,3,5,7,9,11\}$. This set contains an arithmetic progression of length 6, but can we also say it contains arithmetic progressions of length 5, 4, and 3?

In other words, is the statement "a set of natural numbers does not contain arbitrarily long arithmetic progressions" equivalent to the statement "there exists some natural number $N$ such that the set contains no arithmetic progressions of length greater than $N$"?

2. Mar 12, 2012

### Norwegian

Hi Drag,

I would answer yes to both of your question, but I will not subscribe to your in between phrase "In other words".

As far as I know, the conjecture is not even proved for arithmetic progressions of length 3, so I would start there (and believe me, I have tried, and will probably try again).