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        riddles >> putnam exam (pure math) >> Countable Discontinuities
        
(Message started by: william wu on Apr 21^st, 2004, 9:07pm)

Title: Countable Discontinuities
Post by william wu on Apr 21^st, 2004, 9:07pm

Let f be a monotonic increasing function that maps from the reals to the reals. Prove that f can have at most a countable number of discontinuities.

Title: Re: Countable Discontinuities
Post by Icarus on Apr 22^nd, 2004, 4:16pm

Proof:[hide] Since f is monotonic, f(a-) = sup{f(x) : x<a} and f(a+) = inf{f(x) : x > a}, so both the right and left limits exist at every point. If a < b are two discontinuities of f, then f(a+) < f(b-), since f is increasing. Hence the intervals (f(a-), f(a+)) and (f(b-), f(b+)) are disjoint. Thus every discontinuity of f has a unique interval associated with it, disjoint from the intervals of other discontinuities. But each such interval must contain a rational number. Hence there can only be countably many.[/hide]

Title: Re: Countable Discontinuities
Post by Barukh on Apr 23^rd, 2004, 11:06am

Bravo, Icarus! I liked your argument very much! :D

Title: Re: Countable Discontinuities
Post by Icarus on Apr 23^rd, 2004, 4:16pm

I did leave a minor point out: [hide]Because f is increasing, f(x-) <= f(x) <= f(x+). So if f(x-) = f(x+), they must both equal f(x), and f is thus continuous at x. Hence for any discontinuity a, f(a-) < f(a+), and the interval between them is not empty.[/hide]

Let me add this follow-up challenge: If any arbitrary Real function f possesses only jump discontinuities (ie, f(a-) and f(a+) exist but are not equal), is it possible to have uncountably many of them?

Title: Re: Countable Discontinuities
Post by Eigenray on May 2^nd, 2004, 11:33pm

on 04/23/04 at 16:16:42, Icarus wrote:

Let me add this follow-up challenge: If any arbitrary Real function f possesses only jump discontinuities (ie, f(a-) and f(a+) exist but are not equal), is it possible to have uncountably many of them?

For each x [in] [bbr] define t_x = |f(x⁺)-f(x^-)|.
We show that each S_n := {x | t_x > 1/n} is countable, and so, therefore, is their union, [cup]_n=1^[infty]S_n = {x | t_x > 0}.

Fix n. For each x [in] S_n, the limit f(x^-) exists, so there exists x' < x such that |f(y)-f(x^-)| < 1/2n whenever x' < y < x. For any such y, it follows that |f(y⁺)-f(x^-)| and |f(y^-)-f(x^-)| are both [le] 1/2n, so that t_y = |f(y⁺)-f(y^-)| [le] 1/n, which means y [notin] S_n. It follows that the intervals (x',x) are disjoint for distinct values of x [in] S_n, so there are only countably many (each contains a rational).

Title: Re: Countable Discontinuities
Post by Icarus on May 4^th, 2004, 6:04pm

Excellent proof!