# What is the order of discontinuity of this function?

Consider the function f(x) such that f(x) = 0 for all rational x and f(x) = 1 for all irrational x. It would seem that the number of 'jumps' up is uncountably infinite and the number of 'jumps' down is countably infinite; or is the other way around? Shouldn't the number of jumps up have the same order as the number of jumps down? Would the total number of jumps up or down have the same cardinality as the set of all real numbers? Is there a formal 'measure' for the order of discontinuity of a function? Thanks in advance for any guidance you can provide.

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For your fist question: $\Bbb Q$ is countable, while $\Bbb R\setminus \Bbb Q$ is not. – Pedro Tamaroff Jul 19 '12 at 2:44

Normally we say that a function $f$ has a jump at a point $a$ if $\lim\limits_{x\to a^-}f(x)$ and $\lim\limits_{x\to a^+}f(x)$ exist and are not equal. In the case of that function, neither limit exists at any point, so there are no jumps in this sense.
One can instead require that for any sequence $(x_n)_{n\in\mathbb{N}}$ of points in the domain which converges to c, the corresponding sequence $\left(f(x_n)\right)_{n\in \mathbb{N}}$ converges to f(c). In mathematical notation, $\forall (x_n)_{n\in\mathbb{N}} \subset I:\lim_{n\to\infty} x_n=c \Rightarrow \lim_{n\to\infty} f(x_n)=f(c)\,.$
Pick up a sequence $a(n) \in Q$ with a limit point $a \in Q'$ and apply the above definition.