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We've had a number of interesting questions on topological conjugacy on this site. This answer shows how to prove the tent map is topologically conjugate to the quadratic map, and this answer shows that another piecewise linear function can be shown to be topologically conjugate to $F(x) = 4x^3 - 3x$. This got me wondering, given any continuous function, can I find some piecewise linear function that is topologically conjugate to it?

user2944352
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    Hint: Think of a self-map $[0,1]\to [0,1]$ whose fixed-point set is the Cantor set. – Moishe Kohan Apr 17 '22 at 10:06
  • This is an intriguing example, but I worry that I'm not getting the hint. It seems to me that a piecewise linear function could be designed to do this. Though in this case, it would need infinitely many 'pieces' to get the job done? – user2944352 Apr 17 '22 at 11:06
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    Then what is your definition of a piecewise-linear function? – Moishe Kohan Apr 17 '22 at 13:46
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    Thinking about this a little more I see your point. If I let my piecewise-linear function have infinitely many segments the question is trivial since I can approximate any function arbitrarily closely. Otherwise, I can't account for maps with infinitely many fixed points. Do counterexamples with finitely many fixed points also exist? – user2944352 Apr 17 '22 at 17:21
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    Of course: Take a function $f: [0,1]\to [0,1]$ such that $f^{-1}(0)$ is the Cantor set. Such a function can be chosen to have only one fixed point. – Moishe Kohan Apr 18 '22 at 12:16
  • Ah yes! That makes sense. I'll have to think more about this. Thanks for the counterexamples! – user2944352 Apr 19 '22 at 00:45

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