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
. ${A=\{x:a
- $A \neq \emptyset $
- $\exists \delta : a\leq x < a+\delta \Rightarrow x \in A $
- $ b$ is an upper bound and $\exists \delta :b-\delta
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
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?