Proof of the Intermediate Value Theorem?

In summary, a proof of the intermediate value theorem was attempted, but it was found to be incorrect. A new attempt was made using the least upper bound axiom, which was successful. However, after further research, the proof was rewritten using more formal notation and concepts from topology. The proof was found to be correct.
  • #1
jgens
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Does anyone know if the following attempt at a proof would qualify as a rough proof of the intermediate value theorem?

Suppose a function f(x) is continuous on the interval (a,b) where a < c < b and an intermediate value k exists such that f(a) < k < f(b) and lim a→c f(a)=k ⋂ lim b→c f(b)=k. Furthermore, let us suppose that f(c) is not between f(a) and f(b). Based upon an early premise it follows that lim x→c f(x)=k; however, since f(c) is not between f(a) and f(b) this violates our earlier assumption that f(x) is continuous. Therefore, assuming a function f(x) is continuous on the interval (a,b) where a < c < b a value f(c) must exist between f(a) and f(b).

I’m really new to proofs so any suggestions would be helpful. Thanks.
 
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  • #2
Looks to me like you are proving the wrong thing- and something that is not true!

You say "Furthermore, let us suppose that f(c) is not between f(a) and f(b)" and try to arrive at a contradiction.

The intermediate value theorem says that IF k is a number between f(a) and f(b) THEN there exist c between a and b so that f(c)= k.

You are trying to prove "if c is between a and b, then f(c) is between f(a) and f(b)" which is NOT true. Let f(x)= x2, a= -1, b= 2. f(-1)= 1, f(2)= 4, but 0 is between -1 and 2 and f(0)= 0 is NOT between 1 and 2.
 
  • #3
Here's another attempt at a proof, though it's probably not any better (I think it's essentially the same thing).

Suppose a function f(x) is continuous on the interval (a,b) and a number k exists between the upper and lower bound of the set A created by (a,b). k is then a number such that f(a) < k < f(b) and lim a→c f(a)=k ⋂ lim b→c f(b)= k. The previous two limits constitute the right and left hand limits of a function, therefore we may claim that lim x→c f(x)= k. In order for the non-contradiction of the initial premise of continuity f(c)=k and, as a consequence of the right and left hand limits, a < c < b.

Could someone let me know if a modification of this general method will work or if another route is needed to prove the theorem. Again, thanks.
 
  • #4
jgens, this isn't correct. You just assumed the existence of c out of the blue, which defeats the whole attempt at a proof. The IVT will require to use more powerful stuff than what you've shown. You need to use the least upper bound axiom or an equivalent axiom/theorem.
 
  • #5
Thanks Werg22, I'll look into that.
 
  • #6
Having looked into it, I think it requires mathematics considerably more advanced than what I'm aquainted with so I think I'll retire in my attempts to write a proof for the IVT. Thanks for your responses.
 
  • #7
Really? The LUB axiom is quite intuitive. Maybe you looked into a over the top pedantic formulation of it. If you know about limits, then you shouldn't have any difficulty understanding the axiom.
 
  • #8
Well, with guidance and explantion of the Least Upper Bound from Werg22, I was able to write a correct proof of the intermediate value theorem. However, after having looked into more formal notation, I rewrote my proof and am wondering if it is still valid.

Let a function f(x) be continuous on the interval [a,b] and allow for the existence of a number k such that f(a)<k<f(b). Assuming the set created by [a,b] is a non-empty set and x ϵ R, sup{x ϵ [a,b]:f(x)<k}=c. We may then define a set A such that A={x ϵ [a,b]:f(x)<k} and sup{A}=c.

Let us first suppose f(c)<k, then k-f(c)>0. Therefore, by the epsilon delta formulation of a limit 0<|x-c|< δ and k-f(c)>|f(x)-f(c)|>0; thereby resulting with the inequality: k>f(x) on the interval (c-δ,c+δ). However, this violates the property that c is the supremum of A and therefore the relation f(c)<k is not true.

Let us now suppose f(c)>k, then f(c)-k>0. Therefore, by the epsilon delta formulation of a limit 0<|x-c|<δ and f(c)-k>|f(c)-f(x)|>0; thereby resulting with the inequality: f(x)>k on the interval (c-δ,c+δ). However, this too violates the property that c is the supremum of A. Therefore the relation f(c)>k is not true.

Given that neither of the above relations are true and, assuming the continuity of f(x), f(c)=k. Q.E.D.

Thanks.
 
  • #9
The best way to prove the IVT is to prove that the continuous image of a connected set is connected- that's considerably more advanced that basic calculus but once you have the concepts, the proof is simple.
 
  • #10
Halls, I'm terribly sorry if it was implied in your post, but, is the proof above correct? Also, would you - or someone else - mind pointing me towards an explanation of the concepts you posted above?

Thanks.
 
  • #11
jgens, your proof looks ok.

What Halls is talking about uses some concepts from topology:

Let X and Y be topological spaces. A function f: X → Y is continuous if f-1(U) is open in X for every open set U in Y. A separation of X is a pair of nonempty disjoint open subsets of X such that their union is X. X is connected if there is no separation of X.

We're trying to show that if X is connected and f: X → Y is a continuous surjection, then Y is connected too. Suppose Y is not connected; let U and V be a separation of Y. Then f-1(U) and f-1(V) are disjoint sets whose union is X; they are open because f is continuous, and they are nonempty because f is surjective. Thus f-1(U) and f-1(V) constitute a separation of X, so X is not connected.

The tricky part is proving that these definitions of continuity and connectedness are equivalent to the usual analysis definitions.
 
  • #12
I think the following is a quick but maybe primitive proof of the IVT. We want to prove that if f is continuous on an interval [a,b] and if N is any value between f(a) and f(b) then there is a number c on the interval (a,b) such that f(c)=N??
Proof:

Let's consider the following helping function:

g(x)=N-f(x). Now:

g(a)=N-f(a) since f(a)<N<f(b)=> g(a)=N-f(a)>0

Now g(b)=N-f(b)<0 so since g(x) changes sign, then according to the first theorem of Cauhcy the function g(x) must have a root on the interval (a,b) so we have that there is a c in (a,b) such that


g(c)=N-f(c)=0 => f(c)=N, what we wanted to prove.
 
  • #13
Sutupidmath, do you know how the "first theorem of Cauchy" is proven? I'm altogether unfamiliar with the theorem or its proof so I can't really make any assertions, but, it seems to follow as a consequence of the Intermediate Value Theorem. If indeed its proof relies on the IVT, wouldn't you then be assuming the veritability of the IVT in your proof? Maybe I'm just being naive, I don't know.
 
  • #14
jgens said:
Sutupidmath, do you know how the "first theorem of Cauchy" is proven? I'm altogether unfamiliar with the theorem or its proof so I can't really make any assertions, but, it seems to follow as a consequence of the Intermediate Value Theorem. If indeed its proof relies on the IVT, wouldn't you then be assuming the veritability of the IVT in your proof? Maybe I'm just being naive, I don't know.

I believe Cauchy's first theorem is proven independently of the IVT. I'll try to post a proof of it at some later time, sinci i need to take off now.
This theorem basically says, that if a function changes sign at the endpoints of an interval [a,b](i.e. f(a)*f(b)<0 )then there is a point c in that interval such that f(c)=0.
 
  • #15
Well in that case, the IVT is pretty much a trivial consequence of Cauchy's first theorem; the interesting part really would be the proof of that.
 
  • #16
The theorem i was talking about is called i guess: Cauchy's first theorem for continuous functions on a closed interval [a,b]

Theorem: Let f be a continuous functon on the closed interval [a,b]. If f changes sign at the endpoints of the interval (i.e. f(a)f(b)<0) then there exists at least one point c in (a,b) such that f(c)=0.


Proof:

For defeniteness let's suppose that f(a)<0 and f(b)>0. Now let [tex]S=\{x \in [a,b]:f(x)<0\}[/tex].
THat is let S be the set of all points x in [a,b] such that f(x)<0. Now S is clearly nonempty since at least a is in S. Also S is upper bounded by b, so by the Completness Axiom S has a least upper bound, call it c.
Now we wish to prove that actually f(c)=0?
Let's suppose the contrary, that is suppose that f(c) is different from zero, that is f(c)>0 or f(c)<0.

Now, since f is continuous on [a,b] it follows that [tex]\lim_{x\rightarrow c}f(x)=f(c)[/tex]

Now from another theorem, there exists a neighbourhood [tex]\delta>0[/tex] of c such that:

sgn{f(x)}=sgn{f(c)}, for all [tex]x\in (c-\delta,c+\delta)[/tex]

(note: sgn=signum, the sign of ...)

But such a thing is not possible since f(x)<0 for all [tex] x\in(c-\delta, c)[/tex] and

f(x)>0 for all [tex]x\in(c,c+\delta)[/tex]

SO, the contradiction derived means that our assumption that f(c) is different from zero is not correct, thus f(c)=0.


I hope this helps!
 

1. What is the Intermediate Value Theorem?

The Intermediate Value Theorem is a fundamental theorem in calculus that states that if a continuous function f(x) is defined on a closed interval [a, b], and takes on values f(a) and f(b) at the endpoints, then for any value K between f(a) and f(b), there exists a point c in [a, b] such that f(c) = K.

2. What is the significance of the Intermediate Value Theorem?

The Intermediate Value Theorem is significant because it guarantees the existence of solutions or roots to continuous functions. It is used to prove the existence of solutions to equations that cannot be solved algebraically.

3. How is the Intermediate Value Theorem used in real life?

The Intermediate Value Theorem is used in various fields such as economics, engineering, and physics to approximate solutions to problems. For example, it can be used to determine the existence of a temperature at which a chemical reaction will occur, or to find the speed at which an object will hit the ground.

4. Can the Intermediate Value Theorem be applied to all functions?

No, the Intermediate Value Theorem can only be applied to continuous functions. A continuous function is one that does not have any breaks or jumps in its graph, and can be drawn without lifting the pencil from the paper.

5. How is the Intermediate Value Theorem different from the Mean Value Theorem?

While both the Intermediate Value Theorem and the Mean Value Theorem are fundamental theorems in calculus, they have different applications. The Intermediate Value Theorem guarantees the existence of a solution to a continuous function, while the Mean Value Theorem guarantees the existence of a point where the derivative of a function is equal to the average rate of change of the function over a given interval.

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