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 without ever reaching zero?

 

[micromass]

Hi samuelb88!

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(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 $\forall y \in (c-\epsilon, c+\epsilon)$, $f(y) < 0$. 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! :)