1. ## The UnBounded Function

Here is my own problem.

Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.

*I have been able to show that f is discontinous everywhere.

Now I want to find an example of such a function.

* shows how difficult it might be.

Or show it does not exist.

2. Originally Posted by ThePerfectHacker
Here is my own problem.

Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.

*I have been able to show that f is discontinous everywhere.

Now I want to find an example of such a function.

* shows how difficult it might be.

Or show it does not exist.
If it is unbounded on any interval of the real line, how can this really be considered a function?

-Dan

3. Originally Posted by ThePerfectHacker
Here is my own problem.

Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.

*I have been able to show that f is discontinous everywhere.

Now I want to find an example of such a function.

* shows how difficult it might be.

Or show it does not exist.
As we can construct a sequence of nested closed intervals I_n, and our
function is unbounded on all of them it must be infinite (undefined on
(-infty, infnty)) at the common point of the I_n's. This implies that it is undefined
everywhere so does not exist!

RonL

4. Originally Posted by CaptainBlank
As we can construct a sequence of nested closed intervals I_n, and our
function is unbounded on all of them it must be infinite (undefined on
(-infty, infnty)) at the common point of the I_n's. This implies that it is undefined
everywhere so does not exist!

RonL
Did you use the "Nested Interval Theorem"?
If so can you please be very formal, because I never used that theorem except once. Thus, I do not know it well.

5. Originally Posted by ThePerfectHacker
Did you use the "Nested Interval Theorem"?
If so can you please be very formal, because I never used that theorem except once. Thus, I do not know it well.

I've changed my mind the idea that I had in mind does not work.

RonL

1. let f(x)=0 for x irrational

2. Arrange the rationals [0,1) in the order 0, 1/2, 1/3, 2/3, 1/4, 2/4, 3,4,...

3. Then if r_n is the n-th rational in this sequence then put f(r_n)=n

4. Put f(x) = f(frac(x)), frac(x) the fractional part of x.

Conjecture: F as defined above has the property you are looking for
(and if it does not it can be modified slightly to have)

RonL

7. Originally Posted by CaptainBlank

1. let f(x)=0 for x irrational

2. Arrange the rationals [0,1) in the order 0, 1/2, 1/3, 2/3, 1/4, 2/4, 3,4,...

3. Then if r_n is the n-th rational in this sequence then put f(r_n)=n

4. Put f(x) = f(frac(x)), frac(x) the fractional part of x.

Conjecture: F as defined above has the property you are looking for
(and if it does not it can be modified slightly to have)

RonL
I did not check it yet. Because I am about to leave for class.

But, remember the important point I mentioned. IT MUST be discontinous everywhere. It provides an easy check.

Note, if in your function it turns out that f(0)=0 then we have a problem. It is continous at 0.

8. Originally Posted by ThePerfectHacker
I did not check it yet. Because I am about to leave for class.

But, remember the important point I mentioned. IT MUST be discontinous everywhere. It provides an easy check.

Note, if in your function it turns out that f(0)=0 then we have a problem. It is continous at 0.
f(0)=1

but it would not be continuous at 0 even if f(0) did equal 0, as for any n, f(1/n)>n, so there would be points
x arbitarily close to 0 for which f(x) is arbitarily large.

RonL

9. I was discussing this with somebody else today he gave the following construction.

f(x)=0 for irrationals.

f(a/b)=b for rationals where gcd(|a|,b)=1 and a!=0

f(0)=1.

I think that works.

Because small intervals contain rationals with large denominators.

I think we should not forget this function, because it might be useful for disproving certain conjectures about functions.

10. Originally Posted by ThePerfectHacker
I was discussing this with somebody else today he gave the following construction.

f(x)=0 for irrationals.

f(a/b)=b for rationals where gcd(|a|,b)=1 and a!=0

f(0)=1.

I think that works.

Because small intervals contain rationals with large denominators.

I think we should not forget this function, because it might be useful for disproving certain conjectures about functions.
Look at this this is like the sort of thing that I had in mind.

Looking t what I suggested I now note that it needs the refinment (to make
my f single valued):

f(x) = 0 for x irrational.

for x rational f(x) = n, where x=r_n and n is the least n for which
this is true.

RonL