Proof of intermediate value theorem for a particular case

• April 25th 2011, 08:26 AM
poirot
Proof of intermediate value theorem for a particular case
Let
f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that
f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists
a x' such that f(x')=0

S is non empty as x=0 is a member and bounded above by 1, so by the completness axiom Sup(S) exists. Sup(S)_> x E S and Sup(s) _< 1. Not sure where to go from here. Thanks in advance.

• April 25th 2011, 08:47 AM
FernandoRevilla
Denote c = sup S and suppose f ( c ) < 0 or f ( c ) > 0 . In both cases you'll get a contradiction. So, necessarily f ( c ) = 0 .
• April 25th 2011, 08:56 AM
Plato
Quote:

Originally Posted by poirot
[LEFT]Let f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that
f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists a x' such that f(x')=0

This proof relies upon the following theorem.
If $f$ is continuous and http://quicklatex.com/cache3/ql_469a...7f33093_l3.png then there is a neighbored of $c$ on which $f$ has the same sign as $f(c)$.
If you take http://quicklatex.com/cache3/ql_63da...4d8d33f_l3.png and suppose that http://quicklatex.com/cache3/ql_469a...7f33093_l3.png then you can find a contradiction.
• April 25th 2011, 09:14 AM
poirot
let c=Sup(S)
Assume F(c) < 0. I don't see a contradiction
• April 25th 2011, 09:24 AM
Plato
Quote:

Originally Posted by poirot
I don't see it

If http://quicklatex.com/cache3/ql_1f8f...1591d0c_l3.png then http://quicklatex.com/cache3/ql_3f76...b227f26_l3.png.
Can't you find a contradiction there???
• April 25th 2011, 10:05 AM
poirot
I don't know if c is a member of the set. If it is then surely sigma will be 1-c.
• April 26th 2011, 03:05 AM
poirot
ok I'm in a better maths mood today. I got frustated by this yesterday.
assume F(c)<0 which implies the theorem.
c<_1 so F(c)<0 implies c is a member of the set
So for all m E [0,1] with m > c implies that F(m) > 0 but get stuck here. Its sort of intuitive that it can't be because of the density of reals but I don't know how to prove it convincingly.
• April 26th 2011, 04:14 AM
Plato
Quote:

Originally Posted by Plato

It follows that http://quicklatex.com/cache3/ql_91bb...0164acf_l3.png that means http://quicklatex.com/cache3/ql_9212...4fff060_l3.png.
That is a contradiction because c is the greatest number with that property
• April 26th 2011, 04:21 AM
poirot
this is only true if c+ sigma/2 <1
• April 26th 2011, 04:34 AM
Plato
Quote:

Originally Posted by poirot
this is only true if c+ sigma/2 <1

Well of course. But you have to pick delta to make that true.
Look, I am not doing this question for you.
If you do not understand this approach, please do not use it.
• April 26th 2011, 04:51 AM
poirot
sorry but you have put 'there exists a delta >0 '. So you have to show such a delta does not exist, not picking a counter example. Thats just bad logic. I am close to understanding.
• April 26th 2011, 07:37 AM
FernandoRevilla
poirot: I don't understand which is the problem. According to Plato's answer (#5):

http://latex.codecogs.com/png.latex?...,c+%5Cdelta%29

and this, evidently contradicts the choice of c.