# Proof of the Intermediate Value Theorem?

1. Nov 22, 2008

### jgens

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.

Last edited: Nov 22, 2008
2. Nov 23, 2008

### HallsofIvy

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. Nov 23, 2008

### jgens

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. Nov 23, 2008

### Werg22

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. Nov 23, 2008

### jgens

Thanks Werg22, I'll look into that.

6. Nov 23, 2008

### jgens

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. Nov 24, 2008

### Werg22

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. Dec 14, 2008

### jgens

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. Dec 14, 2008

### HallsofIvy

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. Dec 14, 2008

### jgens

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. Dec 14, 2008

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. Dec 15, 2008

### sutupidmath

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. Dec 15, 2008

### jgens

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. Dec 15, 2008

### sutupidmath

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. Dec 15, 2008

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. Dec 15, 2008

### sutupidmath

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 lets suppose that f(a)<0 and f(b)>0. Now let $$S=\{x \in [a,b]:f(x)<0\}$$.
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 $$\lim_{x\rightarrow c}f(x)=f(c)$$

Now from another theorem, there exists a neighbourhood $$\delta>0$$ of c such that:

sgn{f(x)}=sgn{f(c)}, for all $$x\in (c-\delta,c+\delta)$$

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

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

f(x)>0 for all $$x\in(c,c+\delta)$$

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!