The question is as following: Prove the following ‘fixed point theorem’: For every continuous function $f:[0,1]→[0,1]$ there exists $x\in[0,1]$ such that $f(x)=x$

Now I know there is an easy proof using the intermediate value theorem. But I'm trying to prove the same with more preliminary hypotheses. I want to prove it just using completeness and the basic definition of continuous functions. After all, these are the only two things needed to prove the intermediate value theorem, logically it feels like it should be possible to prove the fixed point theorem only with these two. But I'm not able to get past a few lines. I don't know if my approach is right. This is what I have got so far -

Let $ K = \{ x \in [0, 1] : x \le f(x) \}$

$ 0 \in K$ and $K$ is bounded above by $1$, therefore $K$ has a supremum by the Axiom of Completeness

Let $c = Sup \ K$

Now, $f(c) < c$ or $f(c) > c$ or $f(c) = c$

Let if possible $f(c) < c$

Take $\epsilon = c - f(c), \ \exists \ \delta \gt 0$ such that

$|x - c| \lt \delta \Rightarrow |f(x) - f(c)| \lt c - f(c) \Rightarrow f(x) < c$

Taking $x = c + \delta/2$ or $x = c - \delta/2$ isn't getting me anywhere

  • 2
    $\begingroup$ You are already assuming Intermediate Value property in the very first line. Why is the range an interval? $\endgroup$ Jun 11, 2023 at 7:16
  • $\begingroup$ That’s an easy consequence of the brower. $\endgroup$
    – SBF
    Jun 11, 2023 at 7:17
  • 1
    $\begingroup$ I agree with geetha. From wikipedia: In mathematical analysis, the intermediate value theorem states that if $f$ is a continuous function whose domain contains the interval $[a, b],$ then it takes on any given value between $f(a)$ and $f(b)$ at some point within the interval. This has two important corollaries, one of which is: The image of a continuous and not constant function over an interval is itself an interval. So if you're not using IVT to prove the first line in your answer, then how are you proving it? $\endgroup$ Jun 11, 2023 at 7:21
  • 1
    $\begingroup$ I think OP meant "Let the range of $f(x)\subseteq[a,b]$". OP doesn't seem to use that the range equals $[a,b]$, do they? $\endgroup$
    – Al.G.
    Jun 11, 2023 at 8:07
  • $\begingroup$ Thanks for pointing that out, I've edited it out. That anyway wan't playing any role in the approach. $\endgroup$
    – Roshni s
    Jun 11, 2023 at 10:06

1 Answer 1


Hint: your set $K$ is defined by non-strict inequalities of continuous functions ($x\leq f(x)$), hence is closed (for any continuous $g$, the set $\{x:g(x)>0\}$ is obviously open, hence its complement $\{x:g(x)\leq 0\}$ is closed. Now take $g(x)=x-f(x)$ to assert $\{x:x\leq f(x)\}$ closed.)

This means $K$ comes with its suppremum $c$ inside it, that is, $c\leq f(c)$. You only need to disprove $c<f(c)$ now.

To see how to finish I think it's best to draw a picture and see how contradictory it turns out. Then try to explain what is contradictory.

Here's some $f$. On the diagonal I've drawn $(c,c)$ (which we want to show equal to $(c,f(c))$) and we assume that $c<f(c)$:
enter image description here

We want to show $c$ cannot be a supremum if $c<f(c)$, i.e. we need to find points $x>c$ that still have $x\le f(x)$. We'll find them close to the right of $c$ using continuity of $f$.

If $c<f(c)$, then by continuity $f(x)$ is constrained in the blue box $2\delta\times 2\varepsilon$: for $\varepsilon=(f(c)-c)/2$, $$f(c-\delta, c+\delta)\subseteq (f(c)-\varepsilon,f(c)+\varepsilon).$$

Without loss of generality we can shrink (if needed) $\delta<\varepsilon$ so that our blue box does not intersect the diagonal and stays within the upper-left half of the $[0,1]^2$ square.

Now remember that by construction $c$ was the maximum $x$ with $x\le f(x)$. But the entire blue box consists of points strictly above the diagonal line, and within $\delta$ of $c$, we have $f$ constrained exactly there. Hence $c$ is by no means the largest with $x\le f(x)$. Take an arbitrary point to the right, for example $x=c+\delta/2$, then


That is, $c+\delta/2<f(c+\delta/2)$, or $c+\delta/2 \in K$, contradicting supremicity of $c$.

  • $\begingroup$ Thanks you so much! making the box small enough was the key. Taking the epsilon half of f(c) - c and making delta smaller than epsilon cleared it all up. Thanks a lot! $\endgroup$
    – Roshni s
    Jun 11, 2023 at 14:55
  • 1
    $\begingroup$ Glad to help :) $\endgroup$
    – Al.G.
    Jun 11, 2023 at 17:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .