# Assume $f : [0, 1] \rightarrow [0, 1]$ is continuous. Show that there must be a point $x \in [0, 1]$ such that $f(x) = x$ [duplicate]

Assume $f : [0, 1] \rightarrow [0, 1]$ is continuous. Show that there must be a point $x \in [0, 1]$ such that $f(x) = x$

I am not even sure how to begin with this problem, I know that $f$ is continuous so for all $\epsilon > 0$ there is a $\delta > 0$ such that whenever $|x-c|< \delta$ it follows that $|f(x) - f(c)| < \epsilon$

• Consider $g(x)=f(x)-x$, the values of $g(0)$ and $g(1)$, and the Intermediate Value Theorem. – David Mitra Apr 19 '14 at 12:07
• That seems so random, how would I have come up with that? – spitfiredd Apr 19 '14 at 12:13
• @spitfiredd, it is not random at all: when you want to prove that $\;f(x)=g(x)\;$ somewhere under certain conditions, nothing seems to be more natural than looking at the difference function and work with that – DonAntonio Apr 19 '14 at 12:16
• @spitfiredd : $\;f(x)=x\iff f(x)\color{red}-x=0\;$ ...!! – DonAntonio Apr 19 '14 at 12:22
• @spitfiredd , there are lots of tricks...as in may other realms of human life: driving a car, playing the piano, flirting...What can you do to learn as many of these as possible? Practice! – DonAntonio Apr 19 '14 at 12:24

Note that if $f(0)=0$ or $f(1)=1$ we are done immediately, so we can assume $f(0)\gt 0$ $f(1)\lt 1$ and let $y=\sup \left\{x\in [0,1] \land f(x)\ge x\right\}$ - this deploys the completeness of the real numbers: the set is nonempty (contains zero) and is bounded above by $1$.
Then you need to use the continuity of $f$ at $y$ to show that $f(y)=y$.
I put this in to show that you can fall back on the basic concepts when nothing else seems to come to mind. If you follow through the proof you will effectively replicate in this specific case a version of a proof of the intermediate value theorem applied to $f(x)-x$. But it doesn't require you to spot the special function.
It is worth studying the $f(x)-x$ method, though, and understanding how that works, because when you move on to the proofs of mean value theorems and Taylor Series you will see a similar technique in action. It looks more complicated than it is.
The way coming up with the idea is to draw the graph of $y=f(x)$ and the graph of $y=x$. A solution to $f(x)=x$ will be a point where both intersect. But $y=x$ divides the square $[0,1]\times[0,1]$ in half and the graph of $y=f(x)$ must go from one side to the other.
Since we are solving the equation $f(x)=x$, we might as well consider $f(x)-x=0$, which is equivalent, but is also the standard form in which must equations are written in theorems. So, we are trying to prove existence of a zero of $g(x):=f(x)-x$. All theorems that provide existence of solutions of general equations (on the reals) are ultimately equivalent to the intermediate value theorem. This tells us what the hammer should be. Since we also know the right nail is $f(x)-x=0$, it is only a matter of checking the hypotheses of the intermediate value theorem.