# proof of Bolzano’s theorem

Consider the compact interval $[a,b],a and a continuous  real valued function  $f$. If $f(a).f(b)<0$ then there exists $c\in(a,b)$ such that $f(c)=0$

WLOG consider $f(a)<0$ and $f(b)>0$. The other case can be proved using $-f(x)$ which will also verify the theorem’s conditions.

consider $a_{1}=\frac{a+b}{2}$, three cases can occur:

• $f(a_{1})=0$, in this case the theorem is proved $c=a_{1}$

• $f(a_{1})>0$, in this case consider the interval $I_{1}=(a,a_{1})$

• $f(a_{1})<0$, in this case consider the interval $I_{1}=(a_{1},b)$

so starting with an open interval $I_{0}=(a,b)$ we get another open interval $I_{1}\subset I_{0}$ with length half of the original $|I_{1}|=\frac{|I_{0}|}{2}$.

Repeat the procedure to the interval $I_{n}$ and get another interval $I_{n+1}$.

We can thus define a succession of open intervals $I_{n}$ such that $I_{n+1}\subset I_{n}$, $|I_{n}|=2^{-n}|I_{0}|$, such that $I_{n}=(a_{n},b_{n})$ and $f(a_{n})<0.

The succession $c_{2n}=a_{n},c_{2n+1}=b_{n}$ is Cauchy by construction since $m>n\implies|c_{m}-c_{n}|<2^{-[n/2]}|I_{0}|$.

$c_{n}$ is therefore convergent $c_{n}\to c\in[a,b]$, and since $a_{n}$ and $b_{n}$ are sub-successions, they converge to the same limit.

$f$ is continuous in $[a,b]$ so $x_{n}\to x\implies f(x_{n})\to f(x)$

By construction

$f(a_{n})<0$ and $f(b_{n})>0$ so in the limit $\lim_{n\to\infty}f(a_{n})=f(\lim_{n\to\infty}a_{n})=f(c)\leq 0$ and $\lim_{n\to\infty}f(b_{n})=f(c)\geq 0$.

So there exists $c\in[a,b]$ such that $0\leq f(c)\leq 0\implies f(c)=0$.

But since $f(a).f(b)<0$, neither $f(a)=0$ nor $f(b)=0$ and since $f(c)=0$, $c\in(a,b)$

Title proof of Bolzano’s theorem ProofOfBolzanosTheorem 2013-03-22 15:43:29 2013-03-22 15:43:29 cvalente (11260) cvalente (11260) 7 cvalente (11260) Proof msc 26A06