Algebraic structures on the Cantor set

Let $g\in B_{z}^{1}(X)$ and $W\subset B_{z}(X)$ be a neighborhood of the point $g$. Then $g+H(\gamma)\subset W$ for some $\gamma\in\Gamma(X)$. Let $x=r(g)$. Let $x\in V\in\gamma$. The family $\Upsilon(g)$ is finite and $x\notin\bigcup\Upsilon(g)$. Take $U\in{\mathcal{B}}$ such that $x\in U\subset V\setminus\bigcup\Upsilon(g)$.

Let $y\in U$ and $h=g+\{x,y\}$. Then $\Upsilon(g)=\Upsilon(h)$, $R(h)=R(g)+\{x,y\}$ and $r(h)=y$. Let $M=\{g+\{x,y\}\,:\,y\in U\}$. Then $M\subset g+H(\gamma)$ and $r(M)=U$. Hence $r(g)\in U\subset r(W)$.

Let $g\in Y$, $Q=U\cap Y$ and $L=\{g+\{x,y\}\,:\,y\in Q\}$. Then $L=M\cap S$ and $r(L)=Q$. Hence $r(g)\in Q\subset r(W\cap S)$. ∎

Assume that $X$ is a rectifiable space. Let the operations $p$ and $q$ define the structure of a homogeneous algebra, $\Psi$ is a rectification, $\Psi(x,y)=(x,p(x,y))$. We put $\Phi(x,y,z)=q(x,p(y,z))$. $q(x,e)=x$ and $p(x,x)=e$ imply $\Phi(x,y,y)=x$. $q(x,p(x,y))=y$ implies $\Phi(y,y,x)=x$. Let’s check ($\Phi_{3}$). Since $p(y,q(y,p(x,u)))=p(x,u)$ and $q(x,p(x,u))=u$, then

$$\Phi(x,y,\Phi(y,x,u))=q(x,p(y,q(y,p(x,u))))=u.$$

Suppose that $\Phi$ satisfies the conditions ($\Phi_{1}$) and ($\Phi_{3}$). Take $e\in X$ arbitrarily. Let $p(x,y)=\Phi(e,x,y)$ and $q(x,y)=\Phi(x,e,y)$. From ($\Phi_{1}$) follows $p(x,x)=e$ and $q(x,e)=x$. From ($\Phi_{3}$) it follows

	$\displaystyle q(x,p(x,y))$	$\displaystyle=\Phi(x,e,\Phi(e,x,y))=y,$
	$\displaystyle p(x,q(x,y))$	$\displaystyle=\Phi(e,x,\Phi(x,e,y))=y.$

∎

The condition ($\Phi_{3}$) follows from ($\Phi_{1}$) and ($\Phi_{2}$), it suffices to substitute $z=x$ into ($\Phi_{3}$). It remains to apply the Proposition 1. ∎

If the identities ($\Phi_{1}$), ($\Phi_{2}$), and ($\Phi_{3}$) hold on a dense subspace, then they hold everywhere. ∎

We identify $X$ with $\{(x,x)\in Y_{1}\times Y_{2}\,:\,x\in X\}$. Let $Y_{3}$ be the closure of $X$ in $Y_{1}\times Y_{2}$, $\pi_{i}$ be the projection of $Y_{1}\times Y_{2}$ onto $Y_{i}$, $f_{i}=\left.\pi_{i}\right|_{Y_{3}}$ for $i=1.2$, $\Phi_{3}=\left.\Phi_{1}\times\Phi_{2}\right|_{X^{3}}$. Then $\Phi_{3}(X^{3})=X$ and, by the proposition 3, the operations $\Phi_{i}$ for $i=1,2,3$ are a Mal’tsev operation and $f_{i}$ are morphisms with respect to them. Since the spaces $Y_{i}$ are compact, the mappings $f_{i}$ are factorial. Therefore, mappings $f_{i}$ are open (Mal’tsev’s Theorem 4.11 [9], [7]). An open continuous map that is a homeomorphism on a dense set is itself a homeomorphism. Hence the mappings $f_{i}$ are homeomorphisms. Therefore, the mapping $f=f_{1}\circ f_{2}^{-1}:Y_{1}\to Y_{2}$ is a homeomorphism. ∎

We put $\Phi(x,y,z)=xy^{-1}z$. The operation $\Phi$ is the standard Mal’tsev operation on a group. Let’s check ($\Phi_{2}$).

$$\Phi(x,y,\Phi(y,z,u))=xy^{-1}(yz^{-1}u)=xz^{-1}u=\Phi(x,z,u).$$

If $G$ is precompact, then $\Phi$ extends to the completion of $G$. ∎

Let $\tau=\chi(X)$. The compact Mal’tsev space $\widehat{X}$ is a Dugundzhi compact (Theorem 1 [10]). The Dugunji compactum is a dyadic compactum. If a compact dyadic compact contains a dense subspace whose character does not exceed $\tau$, then its weight does not exceed $\tau$ ([11]). ∎

Let $W\subset\widehat{X}$ be open and $W\cap X=U$. Take a non-empty open $V\subset\overline{V}\subset W$. The family $\{\Phi(e,y,V)\,:\,y\in\widehat{X}\}$ forms an open cover of $\widehat{X}$. Then $\bigcup_{i=1}^{n}\Phi(e,x_{i},V)$ for some $x_{1},x_{2},...,x_{n}\in\widehat{X}$. Then $\Phi(e,x_{i},V)\subset\Phi(e,y_{i},U)$ for some $y_{i}\in X$. Let $M=\{y_{1},y_{2},...,y_{n}\}$. ∎

Let $r:B_{z}^{1}({\mathbf{C}})\to{\mathbf{C}}$ as in Theorem 2. Let $\Phi(x,y,z)=r(\{x,y,z\})$. Then $\Phi(x,y,z)\in\{x,y,z\}$ and hence $\Phi$ is a Mal’tsev operation. ∎

There is a compact zero-dimensional extension $bX$ of the space $X$ homeomorphic to ${\mathbf{C}}$, so that for any non-empty open-closed $U,V\subset bX$, $U,V\neq Y$ there exists $f\in\operatorname{Aut}(Y|X)$ so $f(U)=V$, Lemma 2.2 [12]. We will assume that $bX={\mathbf{C}}$.

Denote $\mathcal{C}^{*}=\mathcal{C}\setminus\{{\mathbf{C}}\}$. For $U,V\in\mathcal{C}^{*}$ we set

$$\Theta\left[U,V\right]=f_{U}\circ f_{V}^{-1}:V\to U.$$

It follows from the construction

• ($\Theta_{1}$)

  $\Theta\left[U,V\right]:V\to U$ is a homeomorphism and $\Theta\left[U,V\right](V\cap X)=U\cap X$ for $U,V\in\mathcal{C}^{*}$;

• ($\Theta_{2}$)

  $\Theta\left[U,U\right]=\operatorname{id}_{U}$ for $U\in\mathcal{C}^{*}$;

• ($\Theta_{3}$)

  $\Theta\left[U,V\right]\circ\Theta\left[V,W\right]=\Theta\left[U,W\right]$ for $U,V,W\in\mathcal{C}^{*}$.

Let $x=(x_{n})_{n\in\omega}\in{\mathbf{C}}$, $k\in\omega$. Let’s put

	$\displaystyle U_{k}(x)$	$\displaystyle=\{y\in{\mathbf{C}}\,:\,x|k+1=y|k+1\}=B(x|k+1),$
	$\displaystyle V_{k}(x)$	$\displaystyle=\{y\in{\mathbf{C}}\,:\,x|k=y|k\text{ and }x(k)\neq y(k)\}=U_{k-1}(x)\setminus U_{k}(x)=B(\tilde{x}),$

where $\tilde{x}=(x_{0},...,x_{k-1},1-x_{k})$. Note that

• ($U_{1}$)

  $U_{k}(x),V_{k}(x)\in\mathcal{C}^{*}$;

• ($U_{2}$)

  the family $\{V_{n}(x)\,:\,n\leq k\}$ is an open partition of the set ${\mathbf{C}}\setminus U_{k}(x)$;

• ($U_{3}$)

  the family $\{V_{n}(x)\,:\,{n\in\omega}\}$ is an open partition of the set ${\mathbf{C}}\setminus\{x\}$;

• ($U_{4}$)

  the family $\{U_{n}(x)\,:\,{n\in\omega}\}$ is the base at the point $x$;

• ($U_{5}$)

  $U_{k}(x)=U_{k}(y)$, $V_{k}(x)=V_{k}(y)$ and $x\in U_{k}(y)$ if $y\in U_{k}(x)$;

• ($U_{6}$)

  $V_{k}(x)=U_{k}(y)$, $V_{k}(y)=U_{k}(x)$ and $x\in V_{k}(y)$ if $y\in V_{k}(x)$.

For $x,y\in{\mathbf{C}}$ we define the mapping $h_{x,y}\in\operatorname{Aut}({\mathbf{C}})$. Let $z\in{\mathbf{C}}$. Let’s put

$$h_{x,y}(z)=\begin{cases}x,&z=y\\ \Theta\left[V_{k}(x),V_{k}(y)\right](z),&\text{if $z\in V_{k}(x)$ for some $k\in\omega$}\end{cases}$$

It follows from the construction

• ($h_{1}$)

  $h_{x,y}\in\operatorname{Aut}({\mathbf{C}})$ for $x,y\in{\mathbf{C}}$ and $h_{x,y}\in\operatorname{Aut}({\mathbf{C}}|X)$ for $x,y\in X$;

• ($h_{2}$)

  $h_{x,x}=\operatorname{id}_{\mathbf{C}}$;

• ($h_{3}$)

  $h_{x,y}\circ h_{y,z}=h_{x,z}$;

• ($h_{4}$)

  $h_{x,y}(y)=x$;

• ($h_{5}$)

  $h_{x,y}(U_{k}(y))=U_{k}(x)$ and $h_{x,y}(V_{k}(y))=V_{k}(x)$.

• ($h_{6}$)

  if $x^{\prime}\in U_{k}(x)$ and $y^{\prime}\in U_{k}(y)$ then $\left.h_{x,y}\right|_{V}=\left.h_{x^{\prime},y^{\prime}}\right|_{V}$, where $V=V_{k}(y)$.

Let’s put

$$\Phi:{\mathbf{C}}^{3}\to{\mathbf{C}},\quad(x,y,z)\mapsto h_{x,y}(z).$$

The condition ($\Phi_{1}$) follows from ($h_{2}$) and ($h_{4}$). The condition ($\Phi_{2}$) follows from ($h_{3}$). Hence the operation $\Phi$ is a strong Mal’tsev operation.

Let us prove that $\Phi$ is continuous. Let $x,y,z,u\in{\mathbf{C}}$, $\Phi(x,y,z)=u$, $U$ be a neighborhood of the point $u$. We need to find a neighborhood $W\subset{\mathbf{C}}^{3}$ of the point $(z,y,z)$, so that $\Phi(W)\subset U$.

Consider the case $y=z$. Then $x=u$. Then $S=U_{k}(x)\subset U$ for some $k\in\omega$. Let $V=U_{k}(y)$ and $W=S\times V\times V$. Let’s check $\Phi(W)\subset U$. Let $(x^{\prime},y^{\prime},z^{\prime})\in W$. From ($U_{5}$) and ($h_{5}$) it follows that $h_{x^{\prime},y^{\prime}}(V)=S$. Hence $\Phi(x^{\prime},y^{\prime},z^{\prime})\in U$.

Consider the case $y\neq z$. Then $z\in V_{m}(y)$ for some $m\in\omega$. Let $S=U_{m}(x)$ and $V=U_{m}(y)$. Since the mapping $h_{x,y}$ is continuous, then $h_{x,y}(Q^{\prime})\subset U$ for some neighborhood $Q^{\prime}$ of the point $z$. Let $Q=Q^{\prime}\cap V_{m}(y)$ and $W=S\times V\times Q$. Let’s check $\Phi(W)\subset U$. Let $(x^{\prime},y^{\prime},z^{\prime})\in W$. From ($U_{6}$) it follows that $\left.h_{x^{\prime},y^{\prime}}\right|_{Q}=\left.h_{x,y}\right|_{Q}$. Hence $\Phi(x^{\prime},y^{\prime},z^{\prime})\in U$.

The continuity of $\Phi$ is proved. From ($h_{1}$) it follows that $\Phi(X^{3})=X$. ∎

Let $Y=X\times{\mathbf{D}}$. Then the spaces $X^{\omega}\times{\mathbf{C}}$, $Y^{\omega}$ and $Y^{\omega}\times{\mathbf{D}}$ are homeomorphic. Therefore, $Y^{\omega}$ has a proper open-closed subspace homeomorphic to $Y^{\omega}$. Proposition 24 (5) [13] implies that $Y^{\omega}$ is strongly homogeneous. ∎

Let $Y=X^{\omega}\times{\mathbf{C}}$. Then, by the Proposition 7, $Y$ is strongly homogeneous. The space $X\times Y$ is homeomorphic to $Y$. It remains to apply the Theorem 4. ∎

From the contrary. Take $g\in X\setminus A^{-1}A$. Then $Ag\cap A=\varnothing$. Contradiction. ∎

Assume the contrary, that is, $h({\mathbf{Q}}\times{\mathbf{Q}})\cap{\mathbf{Q}}\times{\mathbf{Q}}=\varnothing$ for some homeomorphism $h:{\mathbf{T}}\to{\mathbf{T}}$. Let $F={\mathbf{Q}}\times\{0\}$. Then $F$ is closed in ${\mathbf{T}}$ and homeomorphic to ${\mathbf{Q}}$. Hence $h(F)$ is closed in ${\mathbf{P}}\times{\mathbf{P}}$ and homeomorphic to ${\mathbf{Q}}$. Contradiction. ∎