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A natural follow-up question one might ask is if there is a function which is continuous on the rational numbers and discontinuous on the irrational numbers. This turns out to be impossible. The set of discontinuities of any function must be an [[Fσ set|{{math|''F''<sub>σ</sub>}} set]]. If such a function existed, then the irrationals would be an {{math|''F''<sub>σ</sub>}} set. The irrationals would then be the [[countable set|countable]] [[union (set theory)|union]] of [[closed set]]s <math display="inline">\bigcup_{i = 0}^\infty C_i</math>, but since the irrationals do not contain an interval, neither can any of the <math>C_i</math>. Therefore, each of the <math>C_i</math> would be nowhere dense, and the irrationals would be a [[meager set]]. It would follow that the real numbers, being the union of the irrationals and the rationals (which, as a countable set, is evidently meager), would also be a meager set. This would contradict the [[Baire category theorem]]: because the reals form a [[complete metric space]], they form a [[Baire space]], which cannot be meager in itself.
A variant of Thomae's function can be used to show that any {{math|''F''<sub>σ</sub>}} subset of the real numbers can be the set of discontinuities of a function. If <math display="inline"> A = \bigcup_{n=1}^{\infty} F_n</math> is a countable union of closed sets <math> F_n</math>, define
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