There is a famous example of a function continuous on irrational numbers but discontinuous on the rational numbers. My question is the converse of this example. Thanks!
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no. Here is an explanation why https://en.wikipedia.org/wiki/Thomae's_function#Follow-up – Doug M Mar 09 '17 at 00:46
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Great! . Thanks. – DIEGO R. Mar 09 '17 at 01:23
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The set of continuity points of a real function must be a $G_\delta$ set, but the rationals aren't $G_\delta$. – symplectomorphic Mar 09 '17 at 02:03
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And here are proofs that $\mathbb{Q}$ isn't $G_\delta$: http://math.stackexchange.com/questions/69451/how-to-show-that-mathbbq-is-not-g-delta – symplectomorphic Mar 09 '17 at 02:05