Unravelling a Proof: If f is Continuous on [a,b] and f(a) < 0 < f(b)

[ak416]

Hi, I am trying to understand the following proof: If f is continuous on [a,b] and f(a) < 0 < f(b), then there is some number x in [a,b] such that f(x) = 0.

It starts with defining A to be {x: a <= x <= b and f is negative on [a,x]}. It uses the least upper bound property to determine that a least upper bound for A does exist and its between a and b. Call this least upper bound "u". Now suppose f(u) < 0. By a continuity theorem there is a d > 0 st f(x) < 0 for u-d < x < u+d. Now there is some x0 in A which satisfies u-d < x0 < u.
This is where i don't understand. Intuitively it makes sense, but how do you know that there's any x between u-d and u+d or u-d and u. All i know is that any x0 in A must be smaller or equal to u. I am trying to make as little assumptions as possible and it seems as if this statement assumes the intermediate value theorem, and that's basically what its trying to prove.

 

[StatusX]

So you know that, by construction, any x with a≤x<u has f(x)<0. You want to show that f(u)=0. This means eliminating the possibilities that is is either greater or less than 0. It looks like first you want to show it can't be less than 0 by assuming it is and then deriving a contradiction. The way to do this is to start with your (true) claim that, assuming f(u)<0, there exists a d>0 such that f(x)<0 for all u-d<x<u+d. This shows right away that u can't be the least upper bound of A, a contradiction. An analagous argument assuming f(u)>0 leads to the same conclusion.

 

[ak416]

Ya i know the the structure of the proof. Its just this specific statement that i don't understand: " there is some x0 in A which satisfies u-d < x0 < u. " Can you explain to me exactly why that's true.

 

[StatusX]

The proof doesn't require x to be in A, although it can be. All you need is to show that u isn't a least upper bound; that either it isn't an upper bound, or there is an upper bound which is less than it. All that is required is that there is some real x in (u-d,u), which is guaranteed since d is positive.

 

[ak416]

ok as a side question. Here's a theorem that's used to prove this: Suppose f is continuous at a, and f(a) > 0. Then there is a number d > 0 such that f(x) > 0 for all x satisfying [x-a] < d.

Now when reading this theorem, do i interpret "for all x" as for all x in the domain of f which can be arbitrary, or can i simply pick any x I can think of that satisfies the relation [x-a] < d. And in a sense, the nature of the domain is stated by the theorem.

Because if its the second case, then ya, i see why the Intermediate Value proof makes sense.

 

[StatusX]

The function is continuous on [a,b], which means [a,b] is part of its domain. So I don't see where the distinction between those two interpretations would come up in the proof. But, in general, d must be chosen so that f is defined at every point with |x-a|<d. If no such d can be chosen, the function isn't continuous at a.