Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
O.k., so by virtue of the definition it was sort of a non-question. Thank you so much. –  Brian Jul 19 '12 at 4:50
add comment

Use the following definition and see what happens

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.