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...

  • $\begingroup$ Do you know the intermediate value theorem? If it has not been discussed, then this question could possibly seem non-trivial. It may also be called "Bolzano's theorem", which is a precursor, $\endgroup$ – астон вілла олоф мэллбэрг Sep 23 '17 at 1:09
  • $\begingroup$ We went through that in class today, but the epsilon and delta stuff is really confusing me $\endgroup$ – kitkat Sep 23 '17 at 1:10
  • $\begingroup$ If you notice, this problem requires the intermediate value theorem. The proof of that fact may require $\epsilon,\delta$ arguments, but once you know IVT this proof is trivial, not requiring any symbols whatsoever. $\endgroup$ – астон вілла олоф мэллбэрг Sep 23 '17 at 1:11

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.


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.

  • $\begingroup$ Thanks a ton! This actually makes a lot more sense to me! $\endgroup$ – kitkat Sep 23 '17 at 14:38
  • $\begingroup$ Thank you. See, initially the $\epsilon-\delta$ style of proofs will drag you out of your comfort zone, because you may not have seen any such proof styles in high school mathematics. However, much like we all learnt the English alphabet, you will look back at this question in 6 months and wonder how you found it so difficult. So do not worry if you are confused, for the moment, about $\epsilon-\delta$ proofs or applications of the IVT . Just give it a few months and everything will look trivial from then on. $\endgroup$ – астон вілла олоф мэллбэрг Sep 24 '17 at 4:46

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