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In the proof of Intermediate Value Theorem, we first consider the case $f(a)>c$ and $f(b)<c$, and then define the set $S = \{x \in [a,b] : f(x)\geq c\}$, and then consider $\sup S=t$ and then show that $f(t)=c$ (the diagram makes it obvious.)

However in the proof they also consider cases $f(t)<c$. But as $t$ is in $S$, $f(t)\geq c$ by definition, so why do we consider that case? If the supremum is not in the set, then too, after all we show that $f(t+\frac{\epsilon}{2})<0$ so that $t+\frac{\epsilon}{2}$ is an upper bound for $S$, but how's that happening? We did not define the set $S$ in such a way right? What mistake am i making? Thanks for helping.

RobPratt
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2 Answers2

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It is not true that $t = \sup S$ implies that $t \in S.$ You don't actually know that $t$ is in $S$ until you have proved by some other means that $f(t) \geq c$.

Since you have not shown the actual proof or cited your source we can only guess, but it appears that you are looking at a proof by contradiction. If we suppose that $f(t)<c$ then some things follow that (as you notice) cannot be true; therefore it cannot be true that $f(t)<c,$ and therefore $f(t)\geq c.$

David K
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  • Its the same as of this https://math.stackexchange.com/questions/3631752/proof-intermediate-value-theorem-correct#comment7463865_3631752 –  Apr 20 '20 at 02:46
  • We are assuming that $t=\sup S$ whereas $S$ is only defined to be ${x \in [a,b] : f(x)\geq c }$ –  Apr 20 '20 at 02:47
  • Anyway I never said $t$ was not $\sup S.$ The way the proof goes (using the notation in your question, not the other question), you don't just "assume" that $t=\sup S$, the very definition of $t$ is that it is $\sup S.$ It still does not follow that $t\in S$ -- it is easy to show a set that does not contain its own supremum -- so your statement that "as $t$ is in $S$, $f(t)\geq c$ by definition" is simply wrong reasoning. – David K Apr 20 '20 at 02:58
  • Ok thanks for helping! :) –  Apr 20 '20 at 02:59
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Note that $t$ may or may not be in $S$ and this does not have any relevance to the proof.

Assume $f(t) <c$ then there is a neighborhood of $I$ of $t$ in which values of $f$ are less than $c$ and hence $I\cap S=\emptyset$. But by definition of supremum every neighborhood of $t$ must contain points of $S$. Thus contradiction implies that $f(t) \geq c$. Prove in similar manner that $f(t) $ does not exceed $c$.

Paramanand Singh
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