Fixed point theorem for the uncountable power of an interval

Question

Does the Brouwer fixed point theorem holds for the uncountable power $[0,1]^\kappa$ of the interval, $\kappa\geq\aleph_1$ ?

That is, does every continuous endomorphism $[0,1]^\kappa\to [0,1]^\kappa$ necessarily have a fixed point ?

A form of the Brouwer fixed point theorem says that for any two maps $f,g:K\to C$ from a compact $K$ to a contractible $C$ necessarily have a coincidence $x\in X$ such that $f(x)=g(x)$ whenever one of them is surjective and has acyclic fibres. Is there a form of this which covers $[0,1]^\kappa$ for $\kappa\geq\aleph_1$ ? (In (Eilenburg and Montgomery,1946) the precise requirements are: $C$ is an acyclic absolute neighborhood retract, and $K$ is a compact metric space. This does not apply to our case.)

Answer 1

Let $f:[0,1]^\kappa\to[0,1]^\kappa$ be given. For a finite subset $F$ of $\kappa$ let $A_F=\{x\in[0,1]^\kappa: \pi_F(x)=\pi_F(f(x))\}$, where $\pi_F$ is the projection onto $[0,1]^F$. The set $A_F$ is closed, and it is also nonempty: define $f_F:[0,1]^F\to[0,1]^F$ by $f_F(x)=\pi_F(f(x^+))$, where $x^+$ is the point of $[0,1]^\kappa$ that extends $x$ by having all other coordinates zero. The map $f_F$ is continuous and has a fixed point $x$, by Brouwer's theorem. Then $x^+\in A_F$.

Clearly $A_{F\cup G}\subseteq A_F\cap A_G$ whenever $F$ and $G$ are finite, so $\{A_F:F\in[\kappa]^{<\omega}\}$ has the finite intersection property. It follows that $\bigcap\{A_F:F\in[\kappa]^{<\omega}\}\neq\emptyset$; that intersection consists of fixed points of $f$.

This proves the fixed-point property of all infinite powers of $[0,1]$ in one go.

The book Fixed Point Theory by Dugundji and Granas mentions a more general result on page 4: A product of compact Hausdorff spaces has the fixed-point property iff every finite subproduct has it. It gives no proof but the above argument establishes the least trivial direction.

Answer 2

The basic form of Brouwer's fixed point theorem does still hold. Fix an uncountable $\kappa$ and a continuous function $f:[0,1]^\kappa \to [0,1]^\kappa$. For any $X \subseteq \kappa$, let $\pi_X : [0,1]^\kappa \to [0,1]^X$ be the natural projection map. For $i<\kappa$, we'll write $\pi_i$ for $\pi_{\{i\}}$, and we'll generally conflate $[0,1]^{\{i\}}$ with $[0,1]$. First we will need the following basic topological fact:

Fact: For any $i<\kappa$, there is a countable $X_i \subset \kappa$ and a continuous map $g_i : [0,1]^{X_i} \to [0,1]$ such that $\pi_i \circ f = g_i \circ \pi_{X_i}$.

Fix such $X_i$'s and $g_i$'s for all $i<\kappa$. Call a set $Y \subset \kappa$ self-sufficient if $X_i \subseteq Y$ for all $i \in Y$. An easy argument shows that for any countable $Y_0 \subset \kappa$, there is a countable self-sufficent $Y \subset \kappa$ with $Y_0 \subseteq Y$.

For any self-sufficient $Y$, let $g_Y : [0,1]^Y \to [0,1]^Y$ be defined by $g_Y(x)(i) = g_i(\pi_{Y_i}(x))$ for all $i \in Y$. It's immediate that $\pi_Y \circ f = g_Y \circ \pi_Y$ for any such $Y$.

By Brouwer's fixed point theorem for $[0,1]^\omega$, each $g_Y$ has a non-empty set of fixed points $F_Y \subseteq [0,1]^Y$. For each such $Y$, let $G_Y = \pi^{-1}_{Y}(F_Y)$. Note that each $G_Y$ is closed.

It is not hard to see that if $Y \subseteq Y'$ are self-sufficient, then $G_Y \supseteq G_{Y'}$. Therefore we have that the family $\mathcal{G} = \{G_Y: Y~\text{countable and self-sufficient}\}$ has the finite intersection property. (Specifically, for any countable self-sufficient $Y$ and $Y'$, we can find a countable self-sufficient $Y'' \supseteq Y \cup Y'$. We then have that $G_{Y''} \subseteq G_Y \cap G_{Y'}$.) By compactness of $[0,1]^\kappa$, $\bigcap \mathcal{G}$ is non-empty. Unwinding definitions gives that any element of $\bigcap \mathcal{G}$ is a fixed point of $f$.

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Proof of fact. In our context, let a basic open be a set of the form $\prod_{j < \kappa} U_j$ where each $U_j \subseteq [0,1]$ is open and for all but finitely many $j<\kappa$, $U_j = [0,1]$. This is the standard basis of the product topology on $[0,1]^\kappa$. Given a basic open $U = \prod_{j<\kappa} U_j$, write $I(U)$ for the set of $j<\kappa$ for which $U_j \neq [0,1]^\kappa$.

For each rational $r < s$ in $[0,1]$, the sets $(\pi_i \circ f)^{-1}([0,r])$ and $(\pi_i \circ f)^{-1}([s,1])$ are disjoint closed subsets of $[0,1]^\kappa$, so, by compactness, we can find finite collections $U_0,U_1,\dots,U_{n-1}$ and $V_0,V_1,\dots,V_{m-1}$ of basic opens such that

• $(\pi_i \circ f)^{-1}([0,r]) \subseteq \bigcup_{k<n} U_k$,

• $(\pi_i \circ f)^{-1}([s,1]) \subseteq \bigcup_{k<m} V_k$, and

• $\bigcup_{k<n} U_k$ and $\bigcup_{k<m} V_k$ are disjoint.

Let $I_{r,s} = \bigcup_{k<n} I(U_k) \cup \bigcup_{k<m} I(V_k)$ for some particular choice of $U_k$'s and $V_k$'s. Finally let $X_i = \bigcup\{I_{r,s}:r<s~\text{rational in}~[0,1]\}$.

Now I claim that for any $x,y \in [0,1]^\kappa$, if $\pi_i(f(x)) \neq \pi_i(f(y))$, then $\pi_{X_i}(x) \neq \pi_{X_i}(y)$. To see this, assume without loss that $\pi_i(f(x)) < \pi_i(f(y))$ and find rational $r<s$ such that $\pi_i(f(x)) < r < s < \pi_i(f(y))$. We now have that $\pi_{I_{r,s}}(x) \neq \pi_{I_{r,s}}(y)$, whence $\pi_{X_i}(x) \neq \pi_{X_i}(x)$.

So now let $g_i: [0,1]^{X_i} \to [0,1]$ be the unique function satisfying that $\pi_i\circ f = g_i\circ \pi_{X_i}$ (which is guaranteed to exist by the above). Now we just need to show that $g_i$ is continuous. Fix an open interval $W \subseteq [0,1]$. The preimage $g_i^{-1}(W)$ is equal to $\pi_{X_i}((\pi_i\circ f)^{-1}(W))$, which is open since $\pi_i\circ f$ is continuous and $\pi_{X_i}$ is open. $\square$

Answer 3

By collapsing cardinals with forcing, one can derive Brouwer's fixed point theorem for $[0,1]^\kappa$ where $\kappa$ is uncountable from Brouwer's fixed point theorem for $[0,1]^{\aleph_0}$.

Suppose that $(X,\mathcal{T})$ is a topological space. Then in a forcing extension $V[G]$, the set $\mathcal{T}$ will no longer be a topology, but it will generate a topology, so any topological space can be interpreted in a forcing extension. But to better interpret a topological space in a forcing extension, we will need to not only add open sets, but we will also need to add points to be able to adequately interpret a topological space in a forcing extension. For example, if we collapse $2^{\aleph_0}$ to have cardinality $\aleph_0$ and we do not add any of the new real numbers to the topological space $\mathbb{R}$, then $\mathbb{R}$ will become countable and totally disconnected which is not what we want because we want to preserve compactness and at least some form of connectedness. I personally prefer to use point-free topology and Boolean valued models when extending a space (or a frame) to a forcing extension because everything works out much more easily that way (one just has to be comfortable with point-free topology).

Suppose that $L$ is a frame, and $B$ is a complete Boolean algebra. Then the frame coproduct $L\oplus B$ will be a $B$-valued frame, and we can consider $L\oplus B$ to be an element of the Boolean valued universe $V^B$. More generally, if $M$ is a frame that contains the complete Boolean algebra $B$ as a subframe, then $M$ is a $B$-valued frame. Many properties of the frame $L$ are preserved when interpreting $L$ as the $B$-valued frame $L\oplus B$ including the following:

1. If $L$ is a compact frame, then $V^B\models``\text{$L\oplus B$ is compact."}$

2. If $L$ is a regular frame, then $V^B\models``\text{$L\oplus B$ is regular."}$

3. If $L=\Omega([0,1]^\kappa)$ and $B$ collapses $\kappa$ to $\aleph_0$, then $V^{B}\models``\text{The frame $L\oplus B$ is homeomorphic to the Hilbert cube."}$

By Hilbert cube, we simply mean $[0,1]^{\aleph_0}$.

If $X$ is a topological space, then we shall let $\Omega(X)$ denote the frame of all open subsets of $X$.

Proposition: Suppose that $X$ is a compact Hausdorff space. If $V^{B}\models``\text{$\Omega(X)\oplus B$ has the fixed point property"}$, then $X$ also has the fixed point property.

Proof: Let $L=\Omega(X)$. Let $f:X\rightarrow X$ be a continuous mapping. Then $\Omega(f):L\rightarrow L$ is a frame homomorphism that induces a frame homomorphism $g:L\oplus B\rightarrow L\oplus B$. By the fixed point property, there is a frame homomorphism $\phi:L\oplus B\rightarrow B$ where $\phi(b)=b$ for each $b\in B$ but where $\phi=\phi\circ g$. In particular, $\phi(U)=\phi(g(U))=\phi(f^{-1}[U])$ for each open $U\subseteq X$.

Suppose now that $\mathcal{M}$ is an ultrafilter on the Boolean algebra $B$. Then $\phi^{-1}[\mathcal{M}]$ is closed under finite intersections, so $\phi^{-1}[\mathcal{M}]$ generates a filter. I claim that $\bigcap_{U\in\phi^{-1}[\mathcal{M}]}\overline{U}$ has just one point. To see this, we observe that whenever $\mathcal{U}$ is an open cover of $X$, we have $\mathcal{U}\cap\phi^{-1}[\mathcal{M}]\neq\emptyset$ by compactness. If $\bigcap_{U\in\phi^{-1}[\mathcal{M}]}\overline{U}$ contains two distinct points $x,y$, then let $f:X\rightarrow[0,1]$ be a continuous with $f(x)=0,f(y)=1.$ Then either $f^{-1}([0,2/3))\in\phi^{-1}[\mathcal{M}]$ or $f^{-1}((1/3,1])\in\phi^{-1}[\mathcal{M}]$. In the case when $f^{-1}([0,2/3))\in\phi^{-1}[\mathcal{M}]$, we have $y\not\in\overline{f^{-1}([0,2/3))}$ which is a contradiction. We get a similar contradiction when $f^{-1}((1/3,1])\in\phi^{-1}[\mathcal{M}]$. Therefore, $\bigcap_{U\in\phi^{-1}[\mathcal{M}]}\overline{U}=x_0$ for some point $x_0$.

If $U\in\phi^{-1}[\mathcal{M}]$, then since $\phi(f^{-1}[U])=\phi(U)$, we have $f^{-1}[U]\in \phi^{-1}[\mathcal{M}]$ as well. Therefore, $x_0\in\overline{f^{-1}[U]}\subseteq f^{-1}[\overline{U}]$ whenever $\phi(U)\in\mathcal{M}$. Therefore, $f(x_0)\in\overline{U}$ whenever $\phi(U)\in\mathcal{M}$, but this is only possible when $x_0=f(x_0).$ $\square$