Subject: categories: RE: stupid question?

From: "Wendt, Michael - SSMD/DMES" <[email-masked]>
Date: Wed, 29 Mar 2000 16:48:16 -0500

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:[email-masked]]
Sent: March 29,2000 4:08 PM
To: [email-masked]
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.