One of my friends pointed out an interesting application of the Brouwer's fixed-point theorem: You cannot stir a coffee in a mug such that all of the coffee particles have changed their position. Namely there exists one particle that is exactly at the same point from which it started. Does the Brouwer's theorem have something to do with permutations, or are permutations not meaningful to represent such a continuous mappings in this case?
Brouwer’s fixed-point theorem, permutations and coffee
fixed-point-theoremsgeneral-topologymetric-spacesrecreational-mathematicssoft-question
Related Solutions
Let $\Omega=[0,1]$. Fix $\varepsilon\in(0,1]$ and define, for each $x\in[0,1]$, \begin{align*} f(x)&\equiv(1-\varepsilon)x,\\ g(x)&\equiv(1-\varepsilon)x+\varepsilon. \end{align*} Clearly, $\|f-g\|_{\infty}=\varepsilon$. However, the only fixed point of $f$ is $0$ and the only fixed point of $g$ is $1$, which are as far from each other as possible.
ADDED: That said, one can establish a continuity property of sorts that is worth exploring. For some $n\in\mathbb N$, let $\Omega$ be a non-empty, convex, compact subset of $\mathbb R^n$. Let $\mathcal C$ denote the set of continuous functions mapping $\Omega$ into itself. Define a correspondence $\Phi:\mathcal C\rightrightarrows\Omega$ as $$\Phi(f)\equiv\text{set of fixed points of $f$}\quad\text{for each $f\in\mathcal C$}.$$ By Brouwer’s theorem, $\Phi(f)$ is not empty for any $f\in\mathcal C$.
Endowing $\mathcal C$ with the supremum norm $\|\cdot\|_{\infty}$ and $\Omega$ with the Euclidean norm $\|\cdot\|_n$, we can establish the following:
THEOREM: The correspondence $\Phi$ is upper hemicontinuous in the sense that if $O$ is an open subset of $\Omega$, then the “inverse image” $$\Phi^{-1}(O)\equiv\{f\in\mathcal C\,|\,\Phi(f)\subseteq O\}$$ is open in $\mathcal C$.
Proof: For the sake of contradiction, suppose that $\Phi^{-1}(O)$ is not open. Then, one can find some $f\in\Phi^{-1}(O)$ and two sequences $(f_m)_{m\in\mathbb N}$ and $(x_m)_{m\in\mathbb N}$ in $\mathcal C$ and $\Omega$, respectively, such that for each $m\in\mathbb N$,
- $\|f_m-f\|_{\infty}<1/m$;
- $x_m\in\Phi(f_m)$; but
- $x_m\in\Omega\setminus O$.
Since $\Omega\setminus O$ is compact, one can take some subsequence $(x_{m_k})_{k\in\mathbb N}$ converging to some $x\in\Omega\setminus O$. For each $k\in\mathbb N$, the following holds: \begin{align*} \|x-f(x)\|_n&\leq \|x-x_{m_k}\|_n+\|x_{m_k}-f_{m_k}(x_{m_k})\|_n\\ &+\|f_{m_k}(x_{m_k})-f(x_{m_k})\|_n+\|f(x_{m_k})-f(x)\|_n. \end{align*} The first, third, and fourth terms converge to $0$ as $k\to\infty$ because of convergence in $\Omega$, convergence in $\mathcal C$, and continuity, respectively. The second term vanishes because $x_{m_k}$ is a fixed point of $f_{m_k}$ for every $k\in\mathbb N$. It follows that $\|x-f(x)\|_n=0$, that is, $x$ is a fixed point of $f$. Since $f\in\Phi^{-1}(O)$, the conclusion is that $x\in \Phi(f)\subseteq O$, which contradicts $x\in\Omega\setminus O$. $\quad\blacksquare$
The above upper-hemicontinuity property of $\Phi$ can be given an equivalent sequential characterization as follows:
THEOREM: Let
- $(f_m)_{m\in\mathbb N}$ be a sequence in $\mathcal C$ converging to $f\in\mathcal C$; and
- $(x_m)_{m\in\mathbb N}$ a sequence in $\Omega$ converging to $x\in\Omega$; such that
- $x_m$ is a fixed point of $f_m$ for each $m\in\mathbb N$, that is, $x_m\in\Phi(f_m)$.
Then, $x$ is a fixed point of $f$, that is, $x\in\Phi(f)$.
Proof: For any $m\in\mathbb N$, \begin{align*} \|x-f(x)\|_n&\leq\|x-x_m\|_n+\|x_m-f_m(x_m)\|_n\\ &+\|f_m(x_m)-f(x_m)\|_n+\|f(x_m)-f(x)\|_n. \end{align*}
Proceed as before. $\quad\blacksquare$
I think there is a way to make this rigorous. To do so, you should keep in mind that your topology $\mathcal T_0$ is not well-defined. Instead it depends on a parameter, namely a choice of a point $p \in D^2$ (what you denoted $x_0$).
To indicate this, let me append $p$ to the definition: $$\mathcal T_0(p) = \{A \subset X \mid p \in A \quad\text{or}\quad A = \emptyset\} $$
So with this in mind, perhaps there is an equivalence going on: the Brouwer fixed point theorem is equivalent to the statement that for each function $f : D^2 \to D^2$, if $f$ is continuous in the standard topology then there exists $p \in D^2$ such that $f$ is continuous in the topology $\mathcal T_0(p)$.
Now, I would not say there is anything deep going on here, the topology $\mathcal T_0(p)$ has been concocted to make this equivalence almost trivial.
Best Answer
The modern definition of permutation is "a one-on-one correspondence from a set (which could be infinite) to itself". The image of the coffee in the mug suggests a continuous permutation of its atoms or its idealized points.
That said, the Brouwer Fixed-Point Theorem applies even when the continuous map is not a permutation, so I wouldn't say the theorem is about permutations. In the coffee analogy: It still works if you could compress the water into a smaller volume of the mug and if some points magically ended up in the same spot after stirring.