0

Why and how do we come up with sup of something to prove this Intermediate Value Theorem? I understand the proof mechanically by following step by step, but I do not really see how from the statement of IVT :Let :[,]→ℝ be continuous and ()<0 and ()>0, then there exists a ∈(,) such that ()=0, we suddenly define something such as a sup to proof it. Please feel free to give me another proof to the IVT but I have read many of the proofs and the concept in each one seems to be arbitrary in everyone of them.

Here is the proof given by Cancan Is there a short proof for the Intermediate Value Theorem ( I included the link and the proof that he wrote as I got downvoted as someone asked me not to put links in my questions)

Proof:

Define ={:()≤} (⋆), and claim ()=

Assum ∈(−+,+),>0 ( is in the neighborhood of )

By definition of continuity ∀>0,|()−()|<⇒−+()<()<()+

In the following, we will manipulate both side of this inequality and prove the intermediate value theorem by contradiction:

1.If ()>, then ()−>0, so set =()−⇒()>()−=()−(()−)=(use the left side of the inequality ) ⇒∀∈(−+,+),()>, so it means that (−) is the least upper bound of the set {:()≤}, which contradicts with the definition of (also a least upper bound and it's not possible to have 2 least upper bounds at the same time.)

2.If ()<, then −()>0,so set =−()⇒()<()+=()+(−())=(use the right side of the inequality)⇒∀∈(−+,+),()<. This means that there exist > such that ()<, which contradicts with the definition of again. (because is the sup of the set {:()≤})

If you set =0, then it's the answer of your question. Hope this can help.

cmk
  • 12,303
Joseph Rock
  • 255
  • 1
    If you want to construct a point $c$ for which $f(c) = u$, it seems natural to define $c$ to be the largest point $x$ in $[a,b]$ for which $f(x) \leq u$. But we don't know that the set $S = { x \in [a,b] \mid f(x) \leq u }$ has a maximum element, so we must instead define $c$ to be the supremum of $S$. That idea doesn't seem so strange to me. Intuitively, it seems that this point $c$ must satisfy $f(c) = 0$, because otherwise it would seem that $f$ must have an abrupt jump at $c$ which would violate the continuity of the function $f$. – littleO May 03 '20 at 06:35
  • did you mean f'(c)=0 here? I thought f(c)=u, why must f(c)=0 ? thank you for your reply littleO – Joseph Rock May 03 '20 at 06:45
  • also would it work if I set f(x)< u instead? – Joseph Rock May 03 '20 at 06:47
  • Oops I meant $f(c) = u$. – littleO May 03 '20 at 06:47
  • You should try to prove IVT yourself. You could start via contradiction and assume that there is no $c$ for which $f(c) =0$. You will find that the result being obvious self evident does not really help in getting a proof. The result is one of the main theorems of calculus and depends on completeness of reals. You should not expect any proof without completeness. – Paramanand Singh May 03 '20 at 10:08
  • I have discussed a few proof in this post. – Paramanand Singh May 03 '20 at 10:11

2 Answers2

1

For this particular proof, the idea is actually pretty straight forward. Draw the graph of a continuous function $f: [a,b] \to \Bbb{R}$, with $f(a) < 0$, and $f(b) > 0$. The purpose for defining $c:= \sup\{x\in [a,b]| \, \, f(x) \leq 0\}$ is to locate the "largest zero of $f$" on $[a,b]$.

The proof I first learnt instead used $\alpha :=\sup\{x\in [a,b]: \, \, \text{$f$ is negative on $[a,x]$}\}$. In this case, $\alpha$ is the "smallest/first zero" of $f$ on the interval $[a,b]$. Essentially, these proofs are telling you how to find the smallest/largest zero of $f$. I like to think of the purpose of $\sup$ in these proofs as to make the notion of "smallest/largest" precise.

peek-a-boo
  • 55,725
  • 2
  • 45
  • 89
  • so in this case of IVT, I could have somehow used the notion of inf instead of sup? – Joseph Rock May 03 '20 at 06:50
  • 1
    @JosephRock sure, but then you'd have to modify the proof appropriately. But the idea is the same: use sup/inf (of an appropriate set) to locate a candidate point where $f$ vanishes. Then, show by using a continuity and contradiction argument that $f$ does indeed vanish at that point. – peek-a-boo May 03 '20 at 06:52
  • when you say where f vanishes, what do you mean by that, isn't f still defined on the rest of the interval – Joseph Rock May 03 '20 at 06:57
  • 1
    @JosephRock I mean, suppose the set you consider is $A = {x \in [a,b]|, , f(x)< 0}$. If you denote $\alpha = \sup A$, then "$f$ vanishes at $\alpha$" means $f(\alpha) = 0$. It's just words to mean that the function evaluated at a certain point of its domain is zero. – peek-a-boo May 03 '20 at 06:58
  • also just for clarification, don't you mean the "last/ largest zero" instead of " smallest/ first zero" on the interval [a,b] , since its a sup and not inf in your defintion? thanks for your reply – Joseph Rock May 03 '20 at 07:07
  • 1
    @JosephRock If you want to locate the largest zero of $f$ on $[a,b]$, then you need a $\leq$ sign (I edited my answer to fix this mistake). But if you just want to find some point where the function is zero, I don't think it matters (but I haven't thought about this too carefully). lol strict vs weak inequalities are sometimes important to distinguish, while sometimes they make no difference; my advice is always go through the argument carefully and see what's what :) – peek-a-boo May 03 '20 at 07:09
  • 1
    @JosephRock In my answer I have two sets. $A_1 = {x\in [a,b]: , \text{$f$ is negative on $[a,x]$}}$, and also $A_2 = {x\in [a,b]: , , f(x) \leq 0}$. Correspondingly we define $\alpha_1 = \sup A_1$ and $\alpha_2 = \sup A_2$. Sometimes, it is possible that $\alpha_1 = \alpha_2$ (if there is only one point in the interval where the function is zero). Sometimes, $\alpha_1 < \alpha_2$. $\alpha_1$ is the "smallest zero", while $\alpha_2$ is the "largest zero" (even though we're taking a sup for both, notice that we're taking a sup of different sets). – peek-a-boo May 03 '20 at 07:12
  • could you define the inf case for finding the zero in your example? I would define it as :=inf{∈[,]: is positive on [,]} , what do you think? – Joseph Rock May 03 '20 at 07:14
  • @JosephRock there's really too many possibilities which currently I don't feel like considering. If you want to check whether a certain approach you have is correct, draw a few simple continuous functions (like sine or cosine graphs, or a linear function, or a cubic function), and try to graphically see what the corresponding sup/inf are. This is the way to build intuition for sup and infs, and the ideas behind the proofs. But anyway, as stated, the set you're taking the inf of is empty, because $f(a)< 0$ by hypothesis, so the infimum doesn't exist. – peek-a-boo May 03 '20 at 07:17
  • You probably want something like $\alpha = \inf {x \in [a,b]: \text{$f$ is positive on $[x,b]$}}$. – peek-a-boo May 03 '20 at 07:19
  • But I think if it's < or ≤, we always get the largest zero anyway, since it is a sup? – Joseph Rock May 03 '20 at 07:19
  • No, it depends which set you're taking the sup of, and it depends on the function. You could get the smallest or largest zero, depending on which set you consider. These are all good questions, but you need to verify some things by yourself using simple example functions. Just sketch a few functions, and calculate the various sup and infs, and try to see which zero it locates. For example, take the interval $[-100,100]$, and consider various functions like $f(x) = x^2 -1$, or $f(x) = \sin x$ or $f(x) = \cos x$ etc. – peek-a-boo May 03 '20 at 07:22
  • Ok i will try that, however I am a little confused as if f is negative, how can there be a zero at all? as in your definition : :=sup{∈[,]: is negative on [,] – Joseph Rock May 03 '20 at 07:27
  • I always try with examples but realised i get confused and tripped up ... sorry english isn't my first language – Joseph Rock May 03 '20 at 07:28
  • just a last question, don't you mean " largest zero " here : The proof I first learnt instead used :=sup{∈[,]: is negative on [,]}. In this case, is the "smallest/first zero" of on the interval [,] – Joseph Rock May 03 '20 at 07:43
  • @JosephRock No, it is correct as stated. – peek-a-boo May 03 '20 at 07:51
  • ok so u are going from right to left? – Joseph Rock May 03 '20 at 07:54
0

The intermediate value theorem is strictly linked to the completeness of the real numbers. The function $f(x)=x^2$ takes the values $0$ and $4$ over the rationals, but doesn't take the value $2$.

Completeness can be stated in different ways; a common one is with the supremum property: every upper bounded set has a supremum.

Since the IVT is equivalent to the statement

if $f$ is continuous over the interval $[a,b]$, $f(a)<0$ and $f(b)>0$, then there exists $c\in(a,b)$ such that $f(c)=0$

I'll stick to this which is simpler. Why is it equivalent? Because $f(c)=u$ is equivalent to $f(c)-u=0$ as well as to $-f(c)+u=0$.

How do we find a zero? The intuitive idea is that if we start from a point $r$ where $f(r)<0$ and move towards $b$, we must cross the $x$-axis somewhere.

Where? Again, the intuition says that we will go nearer and nearer to the $x$-axis until we cross it. So if we consider $N=\{x\in[a,b]:f(x)\le0\}$, the least upper bound of $N$ is the candidate for the crossing, because such point cannot be beyond any other upper bound of $N$, where the function is, by definition, positive.

Set $c=\sup N$. There are two cases, if $f(c)\ne0$: either $f(c)<0$ or $f(c)>0$.

In the first case, continuity tells us that $f$ is negative in a right neighborhood of $c$: so we find elements of $N$ greater than an upper bound of $N$. Contradiction.

In the second case, continuity tells us that $f$ is positive on a left neighborhood of $c$: so there are upper bounds of $N$ which are less than $c$. Contradiction.

egreg
  • 238,574