So we all know of a function that's continuous on the irrationals but discontinuous on the rationals, but what about this:
Are there functions
continuous on and discontinuous on ? Justify.
I'm quite sure the answer's no, but not entirely sure on how to prove it.
Any help would be much appreciated, thanks!
The answer is no.
There is a theorem that states the points where a function f is continuous is a set.
A set, is a set that is the countable intersection of open sets. So the irrations can be written as , where , with being an enumeration of the rationals.
I'll leave it to you to show that isn't a set.
Oops, thanks for the clarification. However, I still stand by the fact that the function isn't what he is asking about. What he wants a function that is continuous on and discontinuous everywhere else. (The example you gave is not continuous at the so right there it fails)
Once you've shown that the irrationals are a G-delta set, you can prove by contradiction.
Assume the the rationals are a G-delta set (say an intersection of countably many . Since the rationals are dense in the reals, each is dense in the reals. Applying the same argument to the irrationals, we get that the intersection of the rationals and the irrationals (which is ) is a countable intersection of open dense sets. This violates the Baire category theorem, which states that any such intersection is non-empty (since the reals are complete).
And this completes the proof!
How do you do that part?Once you've shown that the irrationals are a G-delta set, you can prove by contradiction.
(sorry but i'm a first year maths student and i've only encountered these 5 mins ago when I read this post! However, they sound really interesting. Plus, I tihnk I may be able to use them to help me with some continuity proofs that i'm not super happy with).