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I read in a textbook, which had seemed to have other dubious errors, that one may construct a monotone function with discontinuities at every point in a countable set $C \subset [a,b]$ by enumerating the points as $c_1, c_2, \dots$ and defining $f(x) = \sum_{c_n < x}2^{-n}$. However, if seems that if we let $[a,b] = [0,1], C = \mathbb{Q} \cap [0,1]$, then $f(x)$ is constant $1$ everywhere except 0, an apparent counterexample.

So my question is: how does one construct a monotonic function which has discontinuities precisely on a countable set $C$? Further, are there any relatively easy-to-visualize constructions?

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    Since $\sum_{n=10}^\infty 2^{-n}$ is so small, you can see almost exactly what your function will look like if you plot it (with Matlab, say) using only the ten first rationals in your enumeration.2011-10-02

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The construction is correct. I’ll use your example as an illustration. Let $\{q_n:n\in\omega\}$ be an enumeration of $\mathbb{Q}\cap [0,1]$, and let $f(x)=\sum\limits_{q_n for $x\in [0,1]$.

First consider what happens at some $q_m$: $\lim\limits_{x\to {q_m}^-}f(x) = \sum_{q_n because as $x$ moves up towards $q_m$, $\{q_n:q_n includes more and more of the rationals less than $q_m$. Thus, $f$ is continuous from the left at $q_m$, but for every $x>q_m$ we have $f(x)=\sum_{q_n so $f$ jumps by at least $2^{-m}$ at $q_m$. In fact $\lim_{x\to {q_m}^+}f(x) = f(q_m)+2^{-m},$ and the jump is exactly $2^{-m}$.

At each irrational $a \in [0,1]$, however, $f$ is easily seen to be continuous: $\lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = \sum_{q_n

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    @Bean: That it does, very definitely. So you get a bonus: since there are $2^\omega$ enumerations of a countably infinite set, you get $2^\omega$ examples for a given set of points of discontinuity for the price of one. :-)2011-10-02