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Intermediate Value Theorem. If ƒ is a continuous function on a closed interval [a, b] and if u is any value between ƒ(a) and ƒ(b), then u = ƒ(c) for some c in [a, b].

I'm trying to understand this theorem from Wikipedia(https://en.wikipedia.org/wiki/Intermediate_value_theorem#Proof), it says that if $A:= \{x \in [a,b]:f(x)≤u\}$ and $s:=supA$ then $f(s)=u$.

I can't understand it. Please help me to prove this theorem.

Edited:I don't understand this part on Wikipedia: enter image description here

Elsa Hejazian
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    We can't understand what it is you can't understand. – AlvinL Dec 03 '20 at 11:05
  • why if we show that "if$A:= {x \in [a,b]:f(x)≤u}$ and $s:=supA$ then $f(s)=u$" prove our theorem? what is the connection with them? – Elsa Hejazian Dec 03 '20 at 11:07
  • Use the definition of the sup to pass to the limit : from the left you get $f(s) \leq u$, and from the right you get $f(s) \geq u$. – TheSilverDoe Dec 03 '20 at 11:08
  • Also, verify that $A\neq \emptyset$. – AlvinL Dec 03 '20 at 11:09
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    It shows that $s$ works as a valid value for $c$ in the statement of the theorem because it lies in the interval $[a,b]$ by construction, and satisfies $u=f(s)$. – Jaap Scherphuis Dec 03 '20 at 11:15
  • Is your link correct? I've visited the page but could not see anything to do with IVT. The proof on https://en.wikipedia.org/wiki/Intermediate_value_theorem#Proof is quite clear to me - can you please go through it and point at the exact step you don't understand? –  Dec 03 '20 at 11:39
  • @TheSilverDoe Can you explain in more detail, please? – Elsa Hejazian Dec 03 '20 at 11:39
  • @StinkingBishop How $f(s)=u$?and $sup{x \in [a,b]:f(x)≤u}$ is so confusing for me. I can't get what does he mean by $supA$ – Elsa Hejazian Dec 03 '20 at 11:43
  • @ElsaHejazian $\sup A$ is the supremum of $A$, i.e. the least upper bound. If $m=\sup A$, this means two things: (a) $m$ is an upper bound, i.e. $m\ge a$ for all $a\in A$, and (b) $m$ is the smallest number with this property, i.e. if for another number $m'$ it is also $m'\ge a$ for all $a\in A$, then $m\le m'$. You have to take on board (as one of the fundamental properties of real numbers) that any nonempty and bounded above subset of $\mathbb R$ has a supremum. –  Dec 03 '20 at 11:46
  • (Cont'd) The intuition here is that the graph of $f$ has to cross the line $y=u$ somewhere, so $A$ is the set of the points on the $x$ axis where $f$ is still "below" that line, and then $s=\sup A$ is the, sort-of, "last" such point before $f$ "goes" over the line. (for $x>s, f(x)>u$). The idea is now to prove that the line is actually crossed at $s$, i.e, that the function cannot "jump" from "below" the line to "above" the line, if it is assumed to be continuous. That's it for the intuition, but please look at the formal proof and check which step you don't understand. –  Dec 03 '20 at 11:49
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    @StinkingBishop I edited the question. – Elsa Hejazian Dec 03 '20 at 12:23

2 Answers2

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Let me comment on the sentences you are saying you don't understand:

By the properties of the supremum, there exists some $a^{*}\in (c-\delta ,c]$ that is contained in $S$, and so $f(c)<f(a^{*})+\varepsilon \leq u+\varepsilon$.

As the supremum is the least upper bound of $S$, if there wasn't such $a^*$ contained in $S$, it would be then possible to find a smaller upper bound. Namely, $c-\delta$ would be a smaller upper bound of $S$, if you assume the whole interval $(c-\delta, c]$ is devoid of elements of $S$.

Now pick such $a^*\in S$, and you will have $f(a^*)\le u$, while simultaneously having $f(c)-\varepsilon\lt f(a^*)\lt f(c)+\varepsilon$ (which they assumed is not just valid for $a^*$ but for any value in $(c-\delta, c+\delta)$). Put those two inequalities together to get the desired inequality $f(c)<f(a^{*})+\varepsilon \leq u+\varepsilon$.

Picking $a^{**}\in (c,c+\delta )$, we know that $a^{**}\not \in S$ because $c$ is the supremum of $S$. This means that $f(c)>f(a^{**})-\varepsilon>u-\varepsilon$.

$a^{**}\not\in S$ because $c=\sup S$ is the upper bound of $S$, i.e. all elements of $S$ are "below" or equal to $c$, while $a^{**}$ has been chosen to be greater than $c$.

Not being in $S$ means that $f(a^{**})>u$, but with $f(c)-\epsilon\lt f(a^{**})\lt f(c)+\epsilon$ (which again is not just valid for $a^{**}$ but for any number in $(c-\delta, c+\delta)$) we can put together those two inequalities and get the desired inequality $f(c)>f(a^{**})-\varepsilon>u-\varepsilon$.

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About the Wikipedia proof quoted in revision 3:

By the $(\varepsilon, \delta)$-definition of limit and continuity, the proof one line above writes

$$f(x) - \varepsilon < f(c) < f(x)+\varepsilon \tag 1$$

for all $x \in (c-\delta, c+\delta)$.

$c$ is chosen to be the smallest number that is greater than or equal to every member of $S$ ($c=\sup S$).

So there are some $a^* \in (c-\delta, c]$ that's "just smaller" than $c$ and also $a^*\in S$. So $f(a^*) \le u$.

By the right hand side of inequality $(1)$, $f(c) < f(a^*) + \epsilon$, and so

$$f(c) < u + \varepsilon$$

Similarly, since $c = \sup S$, for $a^{**}\in (c, c+\delta)$ that's "just larger" than $c$, $a^{**}\not\in S$ and so $f(a^{**}) > u$.

By the left hand side of inequality $(1)$, $f(a^{**}) - \varepsilon < f(c)$, and so

$$u - \varepsilon < f(c)$$

peterwhy
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