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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. $\square$

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  • $\begingroup$ Also suggestions on how to generalise it would be helpful as well. $\endgroup$ Dec 6, 2016 at 1:31
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    $\begingroup$ 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 $[0,1]^{\infty}$. $\endgroup$ Dec 6, 2016 at 1:48
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    $\begingroup$ In particular, $C(0,1)$ is not compact, and the image of a compact space is always compact. $\endgroup$ Dec 6, 2016 at 1:56
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    $\begingroup$ 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. $\endgroup$ Dec 6, 2016 at 2:41
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    $\begingroup$ 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. $\endgroup$ Dec 6, 2016 at 3:03

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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.

We can show a space is not compact by finding a discrete infinite closed subset.

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 $C(0,1)$.


The quick proof (for one dimension) is more direct - it uses the intermediate value theorem.

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