proof of Bolzano’s theorem
Consider the compact interval $$ and a continuous^{} real valued function^{} $f$. If $$ then there exists $c\in (a,b)$ such that $f(c)=0$
WLOG consider $$ 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:

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

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$f({a}_{1})>0$, in this case consider the interval ${I}_{1}=(a,{a}_{1})$

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$$, in this case consider the interval ${I}_{1}=({a}_{1},b)$
so starting with an open interval ${I}_{0}=(a,b)$ we get another open interval ${I}_{1}\subset {I}_{0}$ with length half of the original ${I}_{1}=\frac{{I}_{0}}{2}$.
Repeat the procedure to the interval ${I}_{n}$ and get another interval ${I}_{n+1}$.
We can thus define a succession of open intervals ${I}_{n}$ such that ${I}_{n+1}\subset {I}_{n}$, ${I}_{n}={2}^{n}{I}_{0}$, such that ${I}_{n}=({a}_{n},{b}_{n})$ and $$.
The succession ${c}_{2n}={a}_{n},{c}_{2n+1}={b}_{n}$ is Cauchy by construction since $$.
${c}_{n}$ is therefore convergent ${c}_{n}\to c\in [a,b]$, and since ${a}_{n}$ and ${b}_{n}$ are subsuccessions, they converge to the same limit.
$f$ is continuous in $[a,b]$ so ${x}_{n}\to x\u27f9f({x}_{n})\to f(x)$
By construction
$$ and $f({b}_{n})>0$ so in the limit ${lim}_{n\to \mathrm{\infty}}f({a}_{n})=f({lim}_{n\to \mathrm{\infty}}{a}_{n})=f(c)\le 0$ and ${lim}_{n\to \mathrm{\infty}}f({b}_{n})=f(c)\ge 0$.
So there exists $c\in [a,b]$ such that $0\le f(c)\le 0\u27f9f(c)=0$.
But since $$, neither $f(a)=0$ nor $f(b)=0$ and since $f(c)=0$, $c\in (a,b)$
Title  proof of Bolzano’s theorem 

Canonical name  ProofOfBolzanosTheorem 
Date of creation  20130322 15:43:29 
Last modified on  20130322 15:43:29 
Owner  cvalente (11260) 
Last modified by  cvalente (11260) 
Numerical id  7 
Author  cvalente (11260) 
Entry type  Proof 
Classification  msc 26A06 