* stupid question?
@ 2000-03-29 17:36 M.M. Mawanda
0 siblings, 0 replies; 4+ messages in thread
From: M.M. Mawanda @ 2000-03-29 17:36 UTC (permalink / raw)
To: cat-dist
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.
^ permalink raw reply [flat|nested] 4+ messages in thread
* Re: stupid question?
2000-03-29 23:13 Max Kanovitch
@ 2000-03-30 3:00 ` Todd Wilson
0 siblings, 0 replies; 4+ messages in thread
From: Todd Wilson @ 2000-03-30 3:00 UTC (permalink / raw)
To: categories
On Wed, 29 Mar 2000, Max Kanovitch wrote:
> The real fun is about a function f such that
> f is unbounded in any open interval (c,d), and
> in addition to that: f(x+y) = f(x)+f(y).
Using AC/Zorn's Lemma, we can construct 2^(2^Aleph_0) such functions,
as follows. Let B be a basis for the reals R as a rational vector
space. Clearly, |B| = 2^Aleph_0. For any non-empty proper subset C
of B, let g be its characteristic function (g(x)=1 if x in C, g(x)=0
otherwise) and let f be the unique linear extension of g to R. Then f
is linear, and its graph is dense in R^2, since, if c and d are such
that g(c)=1 and g(d)=0, then f(qc+rd) = q for all rationals q,r, and r
can be varied to make qc+rd as close as desired to any given real.
--
Todd Wilson
Computer Science Department
California State University, Fresno
^ permalink raw reply [flat|nested] 4+ messages in thread
* RE: stupid question?
@ 2000-03-29 21:48 Wendt, Michael - SSMD/DMES
0 siblings, 0 replies; 4+ messages in thread
From: Wendt, Michael - SSMD/DMES @ 2000-03-29 21:48 UTC (permalink / raw)
To: categories
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.
^ permalink raw reply [flat|nested] 4+ messages in thread
* Re: stupid question?
@ 2000-03-29 20:23 Peter Freyd
0 siblings, 0 replies; 4+ messages in thread
From: Peter Freyd @ 2000-03-29 20:23 UTC (permalink / raw)
To: categories; +Cc: mm.mawanda
M.M. Mawanda asks:
>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.
No it is not true. For example, the function defined by:
f(x) = if x is irrational then 0 else
if x = p/q where p and q are co-prime then q.
^ permalink raw reply [flat|nested] 4+ messages in thread
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2000-03-29 17:36 stupid question? M.M. Mawanda
2000-03-29 20:23 Peter Freyd
2000-03-29 21:48 Wendt, Michael - SSMD/DMES
2000-03-29 23:13 Max Kanovitch
2000-03-30 3:00 ` Todd Wilson
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