Let $X$ be a compact metric space and $f:X\to X$ be a homeomorphism. The finite sequence $\{x_n\}_{n=0}^{k}$ is called $\epsilon$- chain if $d(f(x_n), x_{n+1})<\epsilon$ for $n=0, \ldots k-1$. The point $x$ can be chain to $y$, if for every $\epsilon>0$, there is $\epsilon$- chain $\{x_n\}_{n=0}^{k}$ with $x_0=x$ and $x_k=y$. The non-empty subset $P\subseteq X$ is called a chain component of $f$, whenevr for $x,y\in P$ , the point $x$ can be chain to $y$ and $y$ can be chain to $x$.
\bf{Question} Let $X$ be a compact metric space. Some paper claim that each chain component of $f$ can be realized as the closure of a single $\epsilon$-chain for any $\epsilon>0$.
Proof of it, not clear for me. Please help.
Thank you
Let $P$ be a chain component and $\varepsilon>0$.
Since $X$ is compact, it can be covered by finite number of balls $B_1,B_2,\ldots,B_n$ of radius $\varepsilon/2$. Among these balls, some intersect $P$ and some do not. Without loss of generality, assume that $B_1,B_2,\ldots, B_m$ (with $m\leq n$) are the balls that intersect $P$. For $k=1,2,\ldots, m$, let $x_k$ be an arbitrary point in $P\cap B_k$. Observe that every point in $P$ is within $\varepsilon$ distance from $\{x_1,x_2,\ldots,x_m\}$.
Since $P$ is chain-transitive, there is an $\varepsilon$-chain from $x_1$ to $x_2$, an $\varepsilon$-chain from $x_2$ to $x_3$, and so on, and finally an $\varepsilon$-chain from $x_{m-1}$ to $x_m$. Gluing all these chains, we get an $\varepsilon$-chain from $x_1$ to $x_m$ that passes through all the points $x_2,x_3,\ldots,x_{m-1}$. Clearly, this chain is $\varepsilon$-dense in $P$.