Elementary proof of fixed point theorem.

1. Jul 23, 2011

ArcanaNoir

I'd like someone to check this proof out for violations of math law. It seems like hackery to me, but then so does a lot of what my professor says, so maybe it's not. If it's flawed, just tell me why. Please don't suggest other ways to prove this, that's my job this weekend.

Thanks :)

p.s. This is my first post using $\LaTeX$, since my discovery that it was there all along (used to see the raw text).

1. The problem statement, all variables and given/known data
$f: [0,1] \to [0,1]$ and $f$ is continuous $\Rightarrow \exists x \in [0,1] : f(x)=x$

2. Relevant equations
Anything elementary calculus. Trying to keep it as simple as possible.

3. The attempt at a solution

let $g(x)=x$
Then $g(0) \le f(x) \le g(1)$
or, $g(a) \le f(x) \le g(b)$ where $[a,b]=[0,1]$

$$\lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x$$
therefore
$$\lim_{x\to x}{f(x)} = x$$

Since $f(x)$ takes on the same value as $g(x)$ at some point $x$, $f(x)=g(x)=x$ at this point.

So, for some $x \in [0,1], f(x)=x$

Last edited: Jul 23, 2011
2. Jul 23, 2011

LCKurtz

This "proof" is flawed. Comments on each line below.

No point renaming x to something else. This just says 0<f(x)<1
And this just says the same thing.

But a and b are constants. But anyway, x is presumably known to be continuous.

Strange way to write a limit and f(x) = x doesn't follow from anything you have done.
You aren't close. You said no suggestions. Do you want a hint?

3. Jul 23, 2011

ArcanaNoir

No thanks on the hint. I guess it's back to the drawing board. Thanks for the input :)