14
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Is $a_i=3i+sin(i)$ and $b_i=a_i+\frac{1}{2^i}$ an exception? $\endgroup$
    – Angela Pretorius
    Mar 10, 2012 at 14:25

1 Answer 1

Reset to default
2
$\begingroup$

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'.

$\endgroup$
2
  • $\begingroup$ Nice approach. A couple things: (1) The argument for $A_N$ to be closed is backwards: if $m>N$ then the $m\alpha_i$ are in the complement of the given set, which should thus be closed. So I don't think we should replace the open intervals with closed intervals in the first step. (2) At the end, when you say "This is based on blah", it sounds like maybe you mean showing $f(x)\to 0$ is based on blah, or the contradiction is, but I guess (?) you mean the whole argument is based on blah. Could be clearer. Also in that sentence, perhaps you mean "along every arithmetic progression"? $\endgroup$
    – user21467
    Mar 31, 2012 at 15:23
  • $\begingroup$ The article you mention seems to be this: W. M. Priestley, Sets thick and thin, Amer. Math. Monthly 83.8 (Oct., 1976), 648–650, jstor. Theorem 3 there answers the question here; Corollary 5 is the folk theorem. $\endgroup$
    – user21467
    Mar 31, 2012 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .