I know one proof of Bolzano's Theorem, which can be sketched as follows:
Set
-
$f$ a continuous function in $[a,b]$ such that ${f(a)<0<f(b)}$.
-
${A=\{x:a<x<b \text{ and } f <0\in[a,x] \}}$
- $A \neq \emptyset $
- $\exists \delta : a\leq x < a+\delta \Rightarrow x \in A
$ - $ b$ is an upper bound and $\exists \delta :b-\delta <x \leq b $ and $x$ is another upper bound of $A$.
From the previous observations, $A$ has a supremum $\alpha$, from which we show $f(\alpha) =0$ ad absurdum.
Suppose $f(\alpha) <0$. Then
$$\exists \delta : \alpha – \delta <x<\alpha +\delta \Rightarrow f(x) <0 $$
Since $\alpha$ is the l.u.b., $$\exists x_0 : \alpha – \delta <x_0<\alpha$$ or else $\alpha$ wouldn't be the l.u.b.
Then $f<0$ in $[a,x_0]$. But if $\alpha < x_1 < \alpha +\delta$ then $f$ is also negative in $[x_0,x_1]$. Thus $f$ is negative in $[a,x_1]$, so $x_1 \in A$, which can't happen since $\alpha$ was the supremum.
The same procedure is used to rule out $f(\alpha) >0$, from where it is concluded that $f(\alpha) =0$.
My main concerns are:
- Is the theorem necesserailly proven using the lub property of $\mathbb R$? (I suppose so).
- How could another proof be constructed?
Best Answer
You need the least upper bound property of $\mathbb R$, because since $\mathbb R$ is essentially $\mathbb Q$ closed under the operation of "taking l.u.b."'s, to see the necessity of it, you only need to notice that the result is not true over $\mathbb Q$ : take a continuous function $f : \mathbb R \to \mathbb R$ that has a unique zero $\alpha$ where $\alpha \notin \mathbb Q$ and $f(\alpha) = 0$, and then restrict your function to $\mathbb Q$, i.e. consider the same function but now $f : \mathbb Q \to \mathbb R$. Your function is still continuous over $\mathbb Q$, but it has no zeros anymore. Therefore, the l.u.b. property is necessary to do this proof.
I believe this answers your second question as well.
Hope that helps,