Thread: 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.

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

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

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.

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

How about this for an idea:

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

Originally Posted by CaptainBlank

How about this for an idea:

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.

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

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.

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