I'm very suspicious of your ability to enumerate the rationals in increasing order but I'm no expert.
hi!
I'm looking for a strictly increasing function on I = [0, 1] that is continuous only at the irrational numbers in I.
After some thought, I came up with the following arrangement.
Enumerate the rational numbers in I as such: (with n a natural number) such that
Let if (for 1 <= k <= n)
and if x is irrational and
Is the function continuous at x irrational? I think so.
For any Epsilon > 0, choose Delta = so that for any x' satisfying we have .
Is the function discontinuous at x rational? I think that can be shown using similar reasoning.
Am I close? If not, can I modify my function to meet the requirement?
Here is an such a function due to Walter Rudin.
Suppose that is a sequence of points in such that converges. Note that convergence is absolute so the terms can be rearranged.
Now define i.e the rational terms so that .
Then define
Now that function is not quite what you asked for, but is should give you some ideas.
Hi again,
In my class, we were shown the following example of a strictly increasing function on (0, 1) that is continuous at the irrationals and discontinuous at the rationals:
Let be a listing of all rationals in (0, 1) and
I was wondering whether the above could be easily extended to the case of I = [0, 1].
I would need to include the rational end points and in the list.
I think the function would need to be modified because otherwise I would get and
and similarly and
Will the following function work for the [0, 1] case?
Thanks again.