proof of Bolzano’s theorem

Consider the compact interval [a,b],a<b and a continuousMathworldPlanetmath real valued functionMathworldPlanetmath f. If f(a).f(b)<0 then there exists c(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 a1=a+b2, three cases can occur:

  • f(a1)=0, in this case the theorem is proved c=a1

  • f(a1)>0, in this case consider the interval I1=(a,a1)

  • f(a1)<0, in this case consider the interval I1=(a1,b)

so starting with an open interval I0=(a,b) we get another open interval I1I0 with length half of the original |I1|=|I0|2.

Repeat the procedure to the interval In and get another interval In+1.

We can thus define a succession of open intervals In such that In+1In, |In|=2-n|I0|, such that In=(an,bn) and f(an)<0<f(bn).

The succession c2n=an,c2n+1=bn is Cauchy by construction since m>n|cm-cn|<2-[n/2]|I0|.

cn is therefore convergent cnc[a,b], and since an and bn are sub-successions, they converge to the same limit.

f is continuous in [a,b] so xnxf(xn)f(x)

By construction

f(an)<0 and f(bn)>0 so in the limit limnf(an)=f(limnan)=f(c)0 and limnf(bn)=f(c)0.

So there exists c[a,b] such that 0f(c)0f(c)=0.

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

Title proof of Bolzano’s theorem
Canonical name ProofOfBolzanosTheorem
Date of creation 2013-03-22 15:43:29
Last modified on 2013-03-22 15:43:29
Owner cvalente (11260)
Last modified by cvalente (11260)
Numerical id 7
Author cvalente (11260)
Entry type Proof
Classification msc 26A06