Continuity question with epsilon and delta I am very confused about how to prove this statement and would appreciate any help I can get. The question is as follows:
Let $I=[0,1]$, and let $f\colon I\to I$ that is continuous. Prove that there is some $x$ in $I$ such that $f(x)=x$.
We have been using the $\epsilon$ and $\delta$ definition of continuity lately in class and I don't know if that would also apply to this problem. Like I said, I am super confused by this...
 A: Here's another way to illustrate ACTOH's idea.  If $f(0)=0$, then we're done.  Otherwise, $f(0) > 0$.  If $f(1) = 1$, then we're again done.  Otherwise $f(1)<1$.    Now draw the $45^o$ line in the unit square $[0,1] \times [0,1]$.  Since $f(x)$ is continuous on $[0,1]$, it is going to cross that line at least once.  At that point $x=f(x)$.  That point is a fixed point.  Notice that there can be several such points, even a continuum of them.
A: I am assuming that the intermediate value theorem you have studied, must be  somewhat like this : 

Given a continuous function $f : [a,b] \to [c,d]$, for all $e \in [c,d]$ there exists $g \in [a,b]$ such that $f(g) = e$.

Now, in our case, we have a function $f$ from $[0,1] \to [0,1]$. What we do is to create a new function out  of this one, and then apply the  IVT on that.  Note this procedure carefully, you wil see it many times.
Let  $g(x) = f(x) - x$. Then,  what is the range of $g$? Note that the smallest and largest value of both $x$ and $f(x)$ are $0$ and $1$ respectively.Hence, the largest value of the difference is $1$, and the smallest value is $-1$.
Hence, $g : [0,1] \to [-1,1]$. By IVT, since $0 \in [-1,1]$,  we know that there is some $h \in [0,1]$ such that $g(h) = 0$. But what is $g(h)$? It is $f(h)-h$, by  definition! Hence, we get $f(h) -h= 0 \implies f(h)=h$, for some $h \in [0,1]$. 
This proof did not go through any $\epsilon-\delta$, howver the proof the IVT itself goes  through that. If you want an explanation as to how the IVT is proved (how these  $\epsilon$s and $\delta$s work with each other) then I can help you.
