- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnline Submission ㆍMy Manuscript - For Reviewers - For Editors
 Some properties of strong chain transitive maps Commun. Korean Math. Soc. 2019 Vol. 34, No. 3, 951-965 https://doi.org/10.4134/CKMS.c180095Published online July 31, 2019 Ali Barzanouni Hakim Sabzevari University Abstract : Let $f:X\to X$ be a continuous map on a compact metric space $(X, d)$ and for an arbitrary $x\in X$, \begin{equation*} \mathcal{SC}_d(x, f):=\{y\,|\, x \text{ can be strong $d$-chain to } y \}. \end{equation*} We give an example to show that $\mathcal{SC}_d(x, f)$ is dependent on the metric $d$ on $X$ but it is a closed and $f$-invariant set. We prove that if $\mathcal{SC}_d(x, f)\supseteq \Omega(f)$ or $f$ has the asymptotic-average shadowing property, then $\mathcal{SC}_d(x, f)= X$. Also, we show that if $f$ has the shadowing property, then $\limsup_{n\in \mathbb{N}}\{f^n\}= \mathcal{SC}_d(f)$ where $\mathcal{SC}_d(f)= \{(x, y) \,|\, y\in\mathcal{SC}_d(x, f)\}$. For each $n\in\mathbb{N}$, we give an example in which $\mathcal{SCR}_d(f^n)\neq \mathcal{SCR}_d(f).$ In spite of it, we prove that if $f^{-1}:(X, d)\to (X,d)$ is an equicontinuous map, then $\mathcal{SCR}_d(f^n)= \mathcal{SCR}_d(f)$ for all $n\in\mathbb{N}$. Keywords : (strong) chain recurrence, (strong) chain transitive map MSC numbers : Primary 37B20, 37C50, 37B35 Downloads: Full-text PDF   Full-text HTML