Re: Re: stupid question?

Re: Re: stupid question?

[Max Kanovitch]

 Dear M.M. Mawanda,
 
 > >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]?

 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).
 
 > Date: Wed, 29 Mar 2000 15:23:16 -0500 (EST)
 > From: Peter Freyd <pjf@saul.cis.upenn.edu>
 > Subject: categories: Re: stupid question?
 > 
 >  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.
 > 

Re: stupid question?

[Todd Wilson]

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

RE: stupid question?

[Wendt, Michael - SSMD/DMES]

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.     

Re: stupid question?

[Peter Freyd]

 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.

stupid question?

[M.M. Mawanda]

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.