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 $f(c)=0$ , $c\\in(a,b)$
随便看	HilbertParallelotope Hilbert parallelotope Hilberts16thProblemForQuadraticVectorFields HilbertsHotel HilbertsIrreducibilityTheorem HilbertsNullstellensatz HilbertSpace Hilbert space HilbertSpacesAreUniformlyConvexSpaces Hilbert spaces are uniformly convex spaces HilbertsProblems HilbertsSixteenthProblem Hilbertsvarepsilonoperator HilbertSymbol Hilbert symbol HilbertSystem Hilbert system HilbertTheorem90 Hilbert Theorem 90 HilbertWeylTheorem Hilbert-Weyl theorem Hilbert’s 16th problem for quadratic vector fields Hilbert’s hotel Hilbert’s irreducibility theorem Hilbert’s Nullstellensatz