10
$\begingroup$

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?

  • 3
    "then f(x) is constant 1 everywhere except 0" - I don't think this claim is correct. In fact, $f$ will be strictly smaller than 1 everywhere.2011-10-02
  • 0
    What are the summation limits and index on the example?2011-10-02
  • 0
    I see. This depends on the enumeration. Great! Question answered.2011-10-02
  • 1
    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

1 Answers 1

19

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

First consider what happens at some $q_m$: $$\lim\limits_{x\to {q_m}^-}f(x) = \sum_{q_nq_m$ we have $$f(x)=\sum_{q_n

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

  • 0
    Yes. My confusion was in that the enumeration changes what the function looks like.2011-10-02
  • 7
    @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