Every continuous map of a closed interval into itself has a fixed point The Question: Please show this theorem: Let $f: I=[a,b] \rightarrow \mathbb{R}$ be a continuous map such that $f(I) \supset I  $. Then $f$ has a fixed point on I. 
My Attempt: Suppose there is a function, $g(x)=f(x)-x$. Since $f(x)$ is a continuous mapping, this implies that $g(x)$ is also a continuous mapping. If $f(a) = a$ and/or $f(b) = b$, we are done. We are going to investigate the case when $f(a) > a$ and $f(b) <b$. We know that $f(I) \supset I$, at least one element of $f(I)$ is not in $I$ but, all elements of $I$ is in $f(I)$. 
Now from the intermediate value theorem, there exists a $c \epsilon (a,b)$ such that for every r in I, f(c) = r. Since $f(a) > a$ and $f(b) <b$, this implies that $f(a)-a >0$ and $ f(b) -b < 0$. By the immediate value theorem, there exists a c in (a,b) such that $g(c)= f(c)-c =0$. This means that $f(c) =c$. Therefore, there is at least one fixed point on I. 
Am I on the right track on this one? 
 A: You have the right idea, but the second paragraph of your attempt is quite poorly written and inaccurate in spots. Here’s a cleaned-up version of the whole thing.

Let $g(x)=f(x)-x$; since $f(x)$ is continuous, so is $g(x)$. If $f(a)=a$ or $f(b)=b$, we’re done, so assume that $f(a)\ne a$ and $f(b)\ne b$. We’re told that $f[I]\subseteq I$, so we must have $f(a)>a$ and $f(b)<b$, and hence $g(a)>0$ and $g(b)<0$. By the intermediate value theorem there is a $c\in(a,b)$ such that $g(c)=0$, and hence $f(c)=c$. Thus, $f$ has at least one fixed point on $I$.

The one real error is when you write this:

Now from the intermediate value theorem, there exists a $c\in(a,b)$ such that for every $r$ in $I$, $f(c) = r$. 

This is clearly impossible: $f(c)$ is a single real number and cannot be equal to $r$ for every $r\in I$ (unless $a=b$, so that $I$ consists of a single point). Moreover, $f$ is the wrong function – you’re interested in $g$ here – and the intermediate value theorem doesn’t even guarantee that for *each $r\in I$ there is some $c_r\in(a,b)$ such that $g(c_r)=r$. It does, however, guarantee that for each $r\in\big(g(b),g(a)\big)$ there is some $c_r\in(a,b)$ such that $g(c_r)=r$, and that’s what we need here.
