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I recently encountered the following function, defined in terms of the (standard, middle-thirds) Cantor set $\mathcal{C}$, on the domain $[0,1]$: $$f(x) = \min\{y \in \mathcal{C}: y \geq x\}$$ Since the Cantor set is compact, and $[x,1]$ is compact for any $x \in [0,1]$, then $\mathcal{C} \cap [x,1]$ is also compact; and then the function above should be well-defined with the minimum (rather than infimum). I think it should also have the following properties:

  1. It is non-decreasing
  2. It is discontinuous at the Cantor ``end-points'': i.e., numbers that can be written as an integer divided by a power of 3. This is a countably infinite set of points.
  3. It satisfies $f(x) = x$ if and only if $x \in \mathcal{C}$.
  4. Because of (1), it has at most countably many discontinuities.

My question is: is this function continuous for any $x$ that is not a Cantor end-point? And, if so, how do I square that with my geometric intuition that the function should "jump up" (i.e, have a discontinuity) any time it reaches a point that is not in the Cantor set? (The meta-question, of course, is whether my construction is invalid or any of my 1-4 are incorrect)

Thank you so much in advance!

Pavel S.
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    Alternatively, if the goal is to show $(4),$ then you don't need to know where it is discontinuous. $f$ is easily show to be monotone, so it can only have countably many discontinuities. (There is a theorem saying monotone functions have countably many discontinuities.) – Thomas Andrews Jul 28 '23 at 21:07
  • I mean, your argument, as written, assumes those endpoints are all of the discontinuities. This is true, but requires a proof, as the rest of your question indicates you don't know if it true. – Thomas Andrews Jul 28 '23 at 21:12
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    how do I square that with my geometric intuition that the function should "jump up" (i.e, have a discontinuity) any time it reaches a point that is not in the Cantor set --- I think what happens is that the jumps at the endpoints of the Cantor set "damp out" as you zoom in on any non-endpoint of the Cantor set, somewhat like what happens with the continuity of Thomae's function at irrational numbers (see also Proof of continuity of Thomae Function at irrationals). – Dave L. Renfro Jul 28 '23 at 22:04
  • Thank you! Thomas -- I understand that all (1)-(4) says is that the endpoints are a subset of the discontinuities. I want to know if there are any more. Dave -- I will take a look and see if I can figure out the analogy of that proof to what I want. – Pavel S. Jul 28 '23 at 22:29
  • A quick note - (2) is incorrect: Not all numbers that can be written as an integer divided by a power of $3$ is a Cantor endpoint (e.g. $\frac{4}{9}$). But your statement that the function is discontinuous at the Cantor endpoints is correct. – jet457 Jul 29 '23 at 02:04

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