1. ## sups and lim

f is a continous function in (0,1) which maintains lim x-->0+ f(x)=-1, and lim x-->1- f(x)=1
let's define: A={x in (0,1)| f(x)=0}

s=supA. i need to prove that f(s)=0.

well, i know that in oter words i need to prove that s is part of the group A, and is its maximum. But i don;t know how to proceed from that, because
my proffesor isn't satisfied with verbal proof. he wants formal mathematical proof [i saw him proving existence of sup using the definition of the limit (epsilon-delta) and by that contradicted the existence of another upper-bound number which smaller than the one we needed to prove]

2. ## Re: sups and lim

Originally Posted by orir
f is a continous function in (0,1) which maintains lim x-->0+ f(x)=-1, and lim x-->1- f(x)=1
let's define: A={x in (0,1)| f(x)=0}, s=supA. i need to prove that f(s)=0.

There two cases: 1) $\displaystyle s\in A$ and 2) $\displaystyle s\notin A$.

Case 1) is trivial.

In case 2), there must be a sequence $\displaystyle (a_n)\subset A$ such that $\displaystyle (a_n)\to s$.
But $\displaystyle f$ is continuous, so what can we say about $\displaystyle f(a_n)\to~?$

3. ## Re: sups and lim

well, we didn't learn sequences in calculus yet...
is there any other way to solve this without using sequences??

4. ## Re: sups and lim

Originally Posted by orir
well, we didn't learn sequences in calculus yet...
is there any other way to solve this without using sequences??
Yes there are other ways. But they are tedious. I will not do it.

You can. If $\displaystyle f(s)\ne 0$ then $\displaystyle (\exists \epsilon>0)(\forall x\in(s-\epsilon,s+\epsilon)[f(x)\ne 0]$.