I'm very suspicious of your ability to enumerate the rationals in increasing order but I'm no expert.
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 .
Now that function is not quite what you asked for, but is should give you some ideas.
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?