interval sequences which contains infinite items of an arithmetic progression

Here's a problem in real analysis which has bothered me and my friends for several days:

For an arbitrary sequence of intervals $(a_i,b_i)$, $a_i$ and $b_i$ tend to infinity and the intersection of any two intervals is empty, must there be an arithmetic progression such that there are infinite items of the progression lying in the interval sequence?

thank you

I think the answer is yes. First replace the open intervals with closed intervals, contained in the open ones. This makes it harder. Then let $ A_N = \alpha \in (0,1] \; \text{, such that } f(n \alpha) = 0 \forall n > N $ where f is the indicator function of the set. $A_N$ is closed since if $\alpha_i \rightarrow \alpha, m\alpha_i \rightarrow m \alpha$ and since all of the $m\alpha_i$ are in the union of the (closed) intervals, $m\alpha$ is also. Also, $[0,1] = \cup 0, A_i$, if the conjecture is false, so under those circs, by the Baire category thm, one of the $A_i$ has non-empty interior. Once that has been achieved it is straightforward to show $f(x) \rightarrow 0$ . That is a contradiction. This is based on a known thm about cont functions, that if they converge along every sequence, they converge. I saw it in the math monthly roughly 1984 , where it was called a 'folk theorem'.