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In Rudin's Principles of Mathematical Analysis, he first proves that the set of points at which a monotonic function is discontinuous is at most countable (Theorem 4.30). Immediately following this proof, he remarks (Remark 4.31) "It should be noted that the discontinuities of a monotonic function need not be isolated. In fact, given any countable subset E of (a,b), which may even be dense, we can construct a function f, monotonic on (a,b), discontinuous at every point of E, and at no other point of (a,b)."

I don't quite follow the logical flow here. What is the meaning of discontinuities not needing to be isolated? Does it mean that the set of points at which the function is discontinuous may have points that are limits points of the set?

Is there a specific reason he is clarifying this via a remark? Does Theorem 4.30 mislead people to think that the discontinuities would be isolated? If so, why?

Finally, what is the significance of the phrase, "which may even be dense" in the remark? Does being dense have anything to do with having points that aren't isolated points?

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According to the statement you may find a function which is discontinuos precisely in each rational number and continuous in every irrational number. I think the warning does not so much address the idea that this particular theorem may mislead people to think discontinuity points are isolated, rather the idea of continuity and our intuition with regard to the concept may lead to such a misconception.

As for your last question: a dense set does in fact not contain isolated points. An isolated point $x$ of $S\subset X$ has a whole neighbourhood $U$ such that $U\cap S=\{x\}$, and this cannot not be true if $S$ is dense.

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  • $\begingroup$ Does the neighborhood $U$ have to be a subset of $X$? $\endgroup$ – Abdu Magdy Feb 6 '18 at 14:26
  • $\begingroup$ @AbduMagdy If you want to say that $x$ is an isolated point of $S$ in $X$ (and that's the only reasonable interpretation here, since no other space have been mentioned) then yes, it has to be an open subset of $X$.. $\endgroup$ – Thomas Feb 6 '18 at 14:48
  • $\begingroup$ I was just imagining if $U$ might be open in $S$ but not in $X$. $\endgroup$ – Abdu Magdy Feb 6 '18 at 14:59
  • $\begingroup$ @AbduMagdy No, This is a statement about $S$ in $X$. Actually,the isolated point is open is $S$ with respect to the subspace topology. $\endgroup$ – Thomas Feb 6 '18 at 16:17

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