proof of Bolzano’s theorem

Consider the compact interval $[a,b],a<b$ 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 $f(a_{n})<0<f(b_{n})$ .

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 $f(c)=0$ , $c\\in(a,b)$

单词	ProofOfBolzanosTheorem
释义	proof of Bolzano’s theorem Consider the compact interval $[a,b],a<b$ 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 $f(a_{n})<0<f(b_{n})$ . 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 $f(c)=0$ , $c\\in(a,b)$
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