How about the following? f(x) = 0 if x is irrational and f(a/b) = a, where a/b is a fraction in lowest terms Certainly within any open interval, there are rationals of arbitrarily large numerator. For a function that is not bounded above or below, how about: f(x) = 0 if x is irrational f(a/b) = a if b is even f(a/b) = -a if b is odd -----Original Message----- From: M.M. Mawanda [mailto:mm.mawanda@nul.ls] Sent: March 29,2000 4:08 PM To: cat-dist@mta.ca Subject: categories: stupid question? I have been asked the following question: Is it true that any function defined in a real number closed interval [a,b] (there is not a hypothesis of continuity) is bounded in an open subinterval (c,d) of [a,b]? My spontaneous was NO. Unfortunately I cannot find a counter-example to disapproved my answer. Can someone help.