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I am trying to prove "a function with the intermediate value property cannot have simple discontinuities", inspiring myself with this SE reply.

Consider $f$ with the Intermediate Value Property and suppose it has a simple discontinuity. Then, $f(a)$, $r$ and $l$ cannot be all identical (where $r$ is the right-handed limit of $f$ at $a$ and $l$ the left-handed one).

Applying an adapted version of the definition $4.33$ from Rudin PMA (which has a typo actually, see here also) we have:

  • $f((a- \delta , a)) \subseteq (l- \epsilon , l+\epsilon )$
  • $f((a,a+\delta )) \subseteq (r-\epsilon , r+\epsilon )$
  • $f(a) \subseteq (f(a) - \epsilon , f(a)+ \epsilon )$

Taking the union of these $3$ bullets, we have:

$$f((a-\delta , a+\delta)) \subseteq (l-\epsilon , l+\epsilon) \cup (r-\epsilon , r+\epsilon )\cup (f(a)-\epsilon , f(a)+\epsilon )$$

But, from here I cannot reach the same conclusion as in the answer because $f((a- \delta , a+ \delta))$ could very well be a subset of, say, $(l- \epsilon, l+ \epsilon)$ and hence, be connected.

PS: I'm sorry that the $\epsilon$'s and $\delta$'s in the bullets are not very rigorous, but I think the idea is correct.

niobium
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    Simply choose $\epsilon$ less than $|r-l|/4$. (By the way, your usage of "be included in" is wrong: you mean "be a subset of". It is the elements of an interval, such as $l$, that are included in that interval.) – TonyK Apr 26 '23 at 14:24
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    @TonyK your $\epsilon$ does not work. E.g. $l=r.$ – Anne Bauval Apr 26 '23 at 14:33
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    @AnneBauval: Yes, you are right. I was conflating simple discontinuities with jump discontinuities, but they are not the same. – TonyK Apr 26 '23 at 14:38
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    @TonyK https://en.wikipedia.org/wiki/Subset : "$A$ is a subset of $B$ may also be expressed as [....] $A$ is included (or contained) in $B$". – Anne Bauval Apr 26 '23 at 16:03

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Since $ l,r$ are not both equal to $f(a),$ assume wlog $l<f(a).$ Let $\epsilon:=(f(a)-l)/2.$ Then for $\delta>0$ small enough, $$f((a-\delta,a))\subseteq(l-\epsilon,l+\epsilon)$$and$$l+\epsilon<f(a)$$ hence $f$ does not have the IVP on $(a-\delta,a].$

Anne Bauval
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