论文标题
对角线分隔连续的平方
Diagonals separating the square of a continuum
论文作者
论文摘要
如果不能将其作为其两个适当的SubContinua的结合，则公制连续性$ x $是不可分配的。如果对于每对$ p，q \ in r $，则$ x $的子集$ r $ a $ x $连续连接。令$ x^{2} $表示$ x $的笛卡尔广场，$δ$ $ x^{2} $的对角线。在\ cite {ka}中，询问是否要使用continuum $ x $，与弧不同，$ x^{2} \setMinusδ$在且仅当$ x $是分解时才连接。在本文中，我们表明这个问题没有任何含义。为了证明非本质的证明，我们使用cantor设置的合适同态同态的动态属性来构建适当的不可塑性continuum $ x $。
A metric continuum $X$ is indecomposable if it cannot be put as the union of two of its proper subcontinua. A subset $R$ of $X$ is said to be continuumwise connected provided that for each pair of points $p,q\in R$, there exists a subcontinuum $M$ of $X$ such that $\{p,q\}\subset M\subset R$. Let $X^{2}$ denote the Cartesian square of $X$ and $Δ$ the diagonal of $X^{2}$. In \cite{ka} it was asked if for a continuum $X$, distinct from the arc, $X^{2}\setminus Δ$ is continuumwise connected if and only if $X$ is decomposable. In this paper we show that no implication in this question holds. For the proof of the non-necessity, we use the dynamic properties of a suitable homeomorphism of the Cantor set onto itself to construct an appropriate indecomposable continuum $X$.
