# Thread: Help with Continuity Proof

1. ## Help with Continuity Proof

I think I kinda know the direction this proof needs to go but I'm not sure how to word it all. Any help would be appreciated.

Let $f:[0,1]\rightarrow [0,1]$ be continuous. Prove that there is a $c\in [0,1]$, such that $f(c)=c.$ [Hint: consider $g(x)=x-g(x)$

I'm thinking a proof by contradiction might be apprpriate here. Assume that there is no c that works and then find a contradiction by using the hint somehow????

2. For which x in [0,1] g(x)>0, for which x in [0,1] g(x)<0 ?

Now use Intermediate value theorem - Wikipedia, the free encyclopedia

3. Suppose that $f(x)\ne x$ for all $x\in[0,1]$ and consider $\displaystyle g:[0,1]\to\{-1,1\}:\frac{f(x)-x}{|f(x)-x|}$. Since the denominator is non-zero this is continuous and surjective. But, this implies that $\{-1,1\}$ is connected....