Need clarification on proof on I.V.T.

[Samuelb88]

Theorem. Suppose that f(x) is continuous on [a,b], a, b \in \mathbb{R}. Then \exists c, a<c<b such that f(c)=0.

Proof. Let c=\sup\{ x : f(x) < 0 \} and observe that c<0. Now let x_n \rightarrow c as n \rightarrow \infty and observe that x_n < 0. Get by transmission of continuity that f(x_n) \rightarrow f(c) as n \rightarrow \infty. Thus we have f(c) = \lim_{n \rightarrow \infty} f(x_n) \leq 0. Now suppose that f(c) < 0, we will derive a contradiction. Now we know that \exists \epsilon such that \forall y \in (c-\epsilon, c+\epsilon), f(y) < 0. Then either y<c<0 or c<y<0. Suppose that c<y<0, but this would contradict our definition of c. Therefore we conclude that our initial supposition was wrong and therefore f(c)=0, as required.

Here's my question: I think I'm missing something about the definition of c, that is, how can we define c to be the least upper bound of the negative values of a continuous function since we can get as close as we want to zero with out ever reaching zero?

[micromass]

Hi samuelb88! :smile:

Where in earth did you get this "proof"? A lot of it makes no sense and is not true.

Here's my attempt to clean things up

Theorem. Suppose that f(x) is continuous on [a,b], a, b \in \mathbb{R}. If f(a)<0 and f(b)>0 then \exists c, a<c<b such that f(c)=0.

Proof. Let c=\sup\{ x : f(x) < 0 \} and observe that c<b (since f(b)>0). Now let x_n \rightarrow c as n \rightarrow \infty with f(xn)<0. Get by transmission of continuity that f(x_n) \rightarrow f(c) as n \rightarrow \infty. Thus we have f(c) = \lim_{n \rightarrow \infty} f(x_n) \leq 0. Now suppose that f(c) < 0, we will derive a contradiction. Now we know that \exists \epsilon such that \forall y \in (c-\epsilon, c+\epsilon), f(y) < 0. But for y>c, this contradicts our definition of c. Therefore we conclude that our initial supposition was wrong and therefore f(c)=0, as required.

[Samuelb88]

So how can we assign a value to the least upper bound of the set \{ x : f(x) < 0 \}? I'm confused since for any real number, say c, such that c<0, we can always find another real number, say d, such that c<d<0.

[micromass]

So how can we assign a value to the least upper bound of the set \{ x : f(x) < 0 \}? I'm confused since for any real number, say c, such that c<0, we can always find another real number, say d, such that c<d<0.

Yes, but the c<0 thingies really don't make any sense. I see no reason why c must be <0.

Let's do an example: take the function f(x)=x-2 on [-10,10]. Then

\{x~\vert~f(x)<0\}=[-10,2[

then the least upper bound is 2, so our c=2. This doesn't mean that f(c)<0 (and indeed, our example shows that this is not the case). But we can always take the least upper bound.

[Samuelb88]

Ahh I was thinking about it incorrectly. Thank you, micromass! :)