intermediate value theorem

[triclino]

For the following proof of the intermediate value theorem ,which i found in wikipedia:

Proof:
Let S be the set of all x in [a, b] such that f(x) ≤ u. Then S is non-empty since a is an element of S, and S is bounded above by b. Hence, by completeness, the supremum c = sup S exists. That is, c is the lowest number that is greater than or equal to every member of S. We claim that f© = u.
Suppose first that f© > u, then f© − u > 0. Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε whenever | x − c | < δ. Pick ε = f© − u, then | f(x) − f© | < f© − u. But then, f(x) > f© − (f© − u) = u whenever | x − c | < δ (that is, f(x) > u for x in (c − δ, c + δ)). This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption.
Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S. However, it also exceeds the least upper bound c of S. The contradiction nullifies this paragraph's opening assumption, as well.

We deduce that f© = u as stated.

 

 

 

I have the folowing questions:

1)How does the author come to the conclusion that ,c-δ is an upper bound for S ,in this part of the proof:

"This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption"

2) How does the author conclude that ,x = c+δ/2 is contained in S ,since he has not proved that [latex]c+\frac{\delta}{2}\leq b[/latex] in the following part of the proof:
"Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S"

3) Nowhere in the proof proves that : a<c<b

[overtone]

Let S be the set of all x in [a, b] such that f(x) ≤ u. Then S is

non-empty since a is an element of S, and S is bounded above by b.

Hence, by completeness, the supremum c = sup S exists. That is, c is the

lowest number that is greater than or equal to every member of S. We

claim that f© = u.

 

For a=1, b=2, c=2, f(x) = 1/x, u=1, the claim is false.

 

(Assuming all that refers to real numbers and functions of them, closed intervals on the real number line, or reasonably analogous entities).

Edited April 4, 2013 by overtone

[triclino]

For a=1, b=2, c=2, f(x) = 1/x, u=1, the claim is false.

 

(Assuming all that refers to real numbers and functions of them, closed intervals on the real number line, or reasonably analogous entities).

if a=1 then f(a)=1 and b =2 then f(b) =1/2.

 

Hence u must be between : 1 AND 1/2 .

 

Go and read the theorem in wiki ,before you come into wrong conclusions

[pwagen]

Link to the source, for anyone wondering:

http://en.m.wikipedia.org/wiki/Intermediate_value_theorem

[Nehushtan]

I have the folowing questions:

1)How does the author come to the conclusion that ,c-δ is an upper bound for S ,in this part of the proof:

"This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption"

If [latex]c-\delta[/latex] is not an upper bound of [latex]S[/latex] there would be an element [latex]x\in S[/latex] such that [latex]c-\delta<x[/latex]. Now [latex]x\in S[/latex] implies [latex]x\leqslant c[/latex] (because [latex]c[/latex] is an upper bound) and [latex]f(x)\leqslant u[/latex] (definition of the set [latex]S[/latex]). But that part of the proof has shown that [latex]f(x)>u[/latex] for all [latex]x\in(c-\delta,\,c+\delta)[/latex], in particular for all [latex]x\in(c-\delta,\,c][/latex]. Hence [latex]c-\delta[/latex] must be an upper bound.

 

2) How does the author conclude that ,x = c+δ/2 is contained in S ,since he has not proved that [latex]c+\frac{\delta}{2}\leq b[/latex] in the following part of the proof:

"Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S"

Yeah, some details have been skipped here. The assumption here is that [latex]f(c )<u<f(b)[/latex]. Hence [latex]c<b[/latex] so [latex]\delta[/latex] can be taken small enough such that [latex]c+\delta<b[/latex].

 

3) Nowhere in the proof proves that : a<c<b

For all [latex]x\in S[/latex], [latex]a\leqslant x\leqslant b[/latex]. So [latex]b[/latex] is an upper bound of [latex]S[/latex]; since [latex]c[/latex] is the least upper bound, [latex]c\leqslant b[/latex]. Hence [latex]a\leqslant x\leqslant c\leqslant b[/latex], i.e. [latex]a\leqslant c\leqslant b[/latex]. The strictness of the inequality signs follows from the fact that [latex]u=f( c)[/latex] is strictly between [latex]f(a)[/latex] and [latex]f(b)[/latex]. (NB: It is enough to have [latex]a\leqslant c\leqslant b[/latex] for the proof to proceed. When the proof is complete, it will follow that [latex]a<c<b[/latex].) Edited April 5, 2013 by Nehushtan

[triclino]

If [latex]c-\delta[/latex] is not an upper bound of [latex]S[/latex] there would be an element [latex]x\in S[/latex] such that [latex]c-\delta<x[/latex]. Now [latex]x\in S[/latex] implies [latex]x\leqslant c[/latex] (because [latex]c[/latex] is an upper bound) and [latex]f(x)\leqslant u[/latex] (definition of the set [latex]S[/latex]). But that part of the proof has shown that [latex]f(x)>u[/latex] for all [latex]x\in(c-\delta,\,c+\delta)[/latex], in particular for all [latex]x\in(c-\delta,\,c][/latex]. Hence [latex]c-\delta[/latex] must be an upper bound.

 

Yeah, some details have been skipped here. The assumption here is that [latex]f(c )<u<f(b)[/latex]. Hence [latex]c<b[/latex] so [latex]\delta[/latex] can be taken small enough such that [latex]c+\delta<b[/latex].

 

 

For all [latex]x\in S[/latex], [latex]a\leqslant x\leqslant b[/latex]. So [latex]b[/latex] is an upper bound of [latex]S[/latex]; since [latex]c[/latex] is the least upper bound, [latex]c\leqslant b[/latex]. Hence [latex]a\leqslant x\leqslant c\leqslant b[/latex], i.e. [latex]a\leqslant c\leqslant b[/latex]. The strictness of the inequality signs follows from the fact that [latex]u=f( c)[/latex] is strictly between [latex]f(a)[/latex] and [latex]f(b)[/latex]. (NB: It is enough to have [latex]a\leqslant c\leqslant b[/latex] for the proof to proceed. When the proof is complete, it will follow that [latex]a<c<b[/latex].)

 

 

If we assume : [latex] \delta< b-c[/latex] we have to consider what happens if [latex]\delta\geq b-c[/latex]

 

Since from trichotomy law we have:

 

[latex] \delta< b-c \vee \delta\geq b-c[/latex]

[uncool]

For the following proof of the intermediate value theorem ,which i found in wikipedia:

Proof:

Let S be the set of all x in [a, b] such that f(x) ≤ u. Then S is non-empty since a is an element of S, and S is bounded above by b. Hence, by completeness, the supremum c = sup S exists. That is, c is the lowest number that is greater than or equal to every member of S. We claim that f© = u.

Suppose first that f© > u, then f© − u > 0. Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε whenever | x − c | < δ. Pick ε = f© − u, then | f(x) − f© | < f© − u. But then, f(x) > f© − (f© − u) = u whenever | x − c | < δ (that is, f(x) > u for x in (c − δ, c + δ)). This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption.

Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S. However, it also exceeds the least upper bound c of S. The contradiction nullifies this paragraph's opening assumption, as well.

 

We deduce that f© = u as stated.

 

 

 

I have the folowing questions:

1)How does the author come to the conclusion that ,c-δ is an upper bound for S ,in this part of the proof:

"This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption"

We already have that c is an upper bound for S. In other words, for any x in S, x < c. Therefore, if we can prove that for every x in S, x is not in (c - δ, c], then every x in S is less than c - δ, and c - δ is an upper bound for S.

2) How does the author conclude that ,x = c+δ/2 is contained in S ,since he has not proved that [latex]c+\frac{\delta}{2}\leq b[/latex] in the following part of the proof:

"Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S"

When it says that there is such a δ, that means that you can choose the δ. So choose one such that that x is less than b.

3) Nowhere in the proof proves that : a<c<b

If c = a, then f© = f(a), so u = f(a), which we know is false. Similar proof for b.

=Uncool-

[overtone]

 

if a=1 then f(a)=1 and b =2 then f(b) =1/2.

 

Hence u must be between : 1 AND 1/2 .

 

Go and read the theorem in wiki ,before you come into wrong conclusions

I was responding to the posted (quoted) claim, not whatever is in wiki. You had questions, and I thought that some of the difficulty might stem from whatver omission or misprint created the false claim. Sorry to have offended.

[Nehushtan]

If we assume : [latex] \delta< b-c[/latex] we have to consider what happens if [latex]\delta\geq b-c[/latex]

Why? [latex]\delta[/latex] is something we can choose here (due to the continuity of the funciton at [latex]c[/latex]) and we choose it to be less than [latex]b-c[/latex].

[triclino]

Why? [latex]\delta[/latex] is something we can choose here (due to the continuity of the funciton at [latex]c[/latex]) and we choose it to be less than [latex]b-c[/latex].

The value of δ is defined by the price of ε.

 

When we say that for every ε>0 there exists a δ>0 such that .......e.t.c....e.t.c we mean that for every value of ε>0 there corresponds a particular volue of δ.

 

In our case for ε=u-f©>0 there corresponds a particular [latex]\delta >0 [/latex].

 

We do not how much is that δ.

 

However we can investigate with the help of trichotomy law what happens if : [latex] \delta\leq b-c\vee \delta>b-c[/latex].

 

On the other hand there is no theorem,definition ,axiom to support the expression : choose [latex]\delta >b-c[/latex]

[Nehushtan]

The value of δ is defined by the price of ε.

 

When we say that for every ε>0 there exists a δ>0 such that .......e.t.c....e.t.c we mean that for every value of ε>0 there corresponds a particular volue of δ.

That’s the definition of uniform continuity, not continuity. For continuity, [latex]\delta[/latex] depends not only on [latex]\epsilon[/latex] but also on each point at which the function is continuous. The intermediate-value theorem applies to continuous rather than uniformly continuous functions. Edited April 6, 2013 by Nehushtan

[uncool]

The value of δ is defined by the price of ε.

No, it isn't. The value of δ is chosen by us to satisfy a property, and as we can see from the property, if we have a δ1 that satisfies the property, then for any δ2 < δ1, δ2 satisfies the property as well; therefore we can choose δ as small as we want (but still positive).

=Uncool-

[triclino]

That’s the definition of uniform continuity, not continuity. For continuity, [latex]\delta[/latex] depends not only on [latex]\epsilon[/latex] but also on each point at which the function is continuous. The intermediate-value theorem applies to continuous rather than uniformly continuous functions.

Even better ,that fixes the value of δ stronger.

 

No, it isn't. The value of δ is chosen by us to satisfy a property

=Uncool-

By us??

[uncool]

 

 

By us??

In more mathematical words:

 

Where the theorem says "Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε", we can rewrite it as: "Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε and δ < (b - c)/2". We can do that by using the following:

Since f is continuous, there is a δ1 such that |f(x) - f©| < ε for any x with |x - c| < δ1. Then let δ = min(δ1, (b - c)/2). Then for any x with |x - c| < δ, |x - c| < δ1, so |f(x) - f©| < ε; we then also have that δ < (b - c)/2. So there is such a δ - so we can choose such a δ.

=Uncool-

[Nehushtan]

Even better

Did you read what I wrote? I said: The IVT is about continuous functions, not uniformly continuous functions. You are trying to apply the definition of uniformly continuous function to the IVT – which is not what the theorem is about. That’s why you’re not getting anywhere closer towards understanding the proof. Edited April 7, 2013 by Nehushtan

[triclino]

In more mathematical words:

 

Where the theorem says "Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε", we can rewrite it as: "Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε and δ < (b - c)/2". We can do that by using the following:

Since f is continuous, there is a δ1 such that |f(x) - f©| < ε for any x with |x - c| < δ1. Then let δ = min(δ1, (b - c)/2). Then for any x with |x - c| < δ, |x - c| < δ1, so |f(x) - f©| < ε; we then also have that δ < (b - c)/2. So there is such a δ - so we can choose such a δ.

=Uncool-

We know that [latex] a\leq c\leq b[/latex] ,since c is the Sup of S.

 

Hence [latex]b-c\geq 0\Longrightarrow \frac{b-c}{2}\geq 0[/latex]

 

And in the case where b-c/2 >0 we can choose ΄0<δ <b-c/2 .

 

But in the case where b-c/2 =0 ,if we choose δ<b-c/2 , then δ will be negative. But we want δ positive.

 

Did you read what I wrote? I said: The IVT is about continuous functions, not uniformly continuous functions. You are trying to apply the definition of uniformly continuous function to the IVT – which is not what the theorem is about. That’s why you’re not getting anywhere closer towards understanding the proof.

Where in my previous posts there as the slightest indication of that claim

[uncool]

We know that [latex] a\leq c\leq b[/latex] ,since c is the Sup of S.

 

Hence [latex]b-c\geq 0\Longrightarrow \frac{b-c}{2}\geq 0[/latex]

 

And in the case where b-c/2 >0 we can choose ΄0<δ <b-c/2 .

 

But in the case where b-c/2 =0 ,if we choose δ<b-c/2 , then δ will be negative. But we want δ positive.

We know that (b - c)/2 is nonzero by the paragraph before; if (b - c)/2 = 0, then b = c, so f(b) = f© <= u, and we assumed that f(b) > u.

=Uncool-

Edited April 14, 2013 by uncool