# Please verify my proof of the 1 Dimentional Brouwer fixed-point theorem

I wish to prove that all continuous functions $[0,1] \to [0,1]$ must have a fixed point.

$\mathcal{C}(0,1)$ is the set of all continuous functions $[0,1] \to [0,1]$. Since $| \mathbb{R} | = | \mathcal{C}(0,1) |$ and $C(0,1)$ is path connected it is possible to create a continuous surjection $U : [0,1] \to C(0,1)$.

Assume there exists a fixed point free function $f \in \mathcal{C}(0,1)$. We can then define a function $\Delta(x) = f(U(x)(x))$. $\Delta$ will be continuous since continuity is closed under composition.

Since $\Delta \in \mathcal{C}(0,1)$ we then can find a $\Phi_\Delta$ such that $U(\Phi_\Delta) = \Delta$.

By substitution $\Delta(\Phi_\Delta) = f(U(\Phi_\Delta)(\Phi_\Delta)) = f(\Delta(\Phi_\Delta))$ which contradicts $f$ being fixpoint free hence there can not be a fix point free continuous function. $\square$

• Also suggestions on how to generalise it would be helpful as well. Dec 6, 2016 at 1:31
• There is no continuous function on $[0,1]$ which is onto $\mathbb R$, but your argument would apply there. You need something more than hand-waving, you need proof. You can find a surjective map from $[0,1]$ to $[0,1]^n$ for any finite $n$, but $C(0,1)$ is more like $[0,1]^{\infty}$. Dec 6, 2016 at 1:48
• In particular, $C(0,1)$ is not compact, and the image of a compact space is always compact. Dec 6, 2016 at 1:56
• Let $I_{k}$ be the interval $\left(\frac1{k+1},\frac 1k\right)$. Define a function $f_k$ which is zero outside $I_k$ and is $1$ at the midpoint of $I_k$. Then $f_1,f_2,\cdots$ is an infinite closed discrete subset of $C(0,1)$, and hence not compact. But a closed subset of a compact space is compact. So $C(0,1)$ is not compact. Dec 6, 2016 at 2:41
• Was there a comment deleted? That comment doesn't appear to be related to anything. The function $V$ is a composition of the two maps $(U,\mathrm{id}):[0,1]\to C[0,1]\times [0,1]$ and the evaluation map: $e:C[0,1]\times [0,1]\to [0,1]$ sending $(f,x)\mapsto f(x)$. $e$ is continuous. To get that $V$ is continuous, you'll need $U$ continuous. Dec 6, 2016 at 3:03

Your proof does not work, because you cannot find a continuous $U$.
There cannot be a continuous onto map from $[0,1]$ to $C(0,1)$, because $[0,1]$ is compact, and we can show that $C(0,1)$ is not compact.
In this case, let $I_k$ be the interval $\left(\frac1{k+1},\frac 1k\right)$, and define $f_k$ so that it is zero outside $I_k$ and $1$ at the middle of $I_k$. Then the set $\{f_k\}$ is a closed infinite discrete subset of $C(0,1)$.