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Let $f: R \rightarrow R$. Show that the set of points of continuity of $f$ is a $G_{\delta}$ set. Explain why it follows from this that there is no function that is continuous on the rationals and discontinuous on the irrationals.

Solution:

I can prove the second part, although independently of the first.

In words:

No function can be continuous only on a countable dense set of $R$, such as $Q$. If the set, $X$, of continuity points were countable, then we could choose a nested sequence of intervals around points of $X$ where the variation in $f$ goes to 0, that eventually avoids all points of $X$. But the common point of the intervals would be a continuity point, contradiction.

How can I relate this to the first part?

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Related: here, here, and here. –  Zev Chonoles Apr 23 '13 at 22:57

1 Answer 1

When you look at the intersection of a sequence of set of reals where oscillation (which you call variation) of f goes to 0, all you can say is that this is a G-delta set containing all rationals. How do you know this set has an irrational number? The idea is to exploit the fact that rationals are countable to build a nested sequence of compact intervals whose intersection avoids all rationals. This argument is called the Baire category theorem.

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