Latest revision as of 03:18, 26 April 2015

The set of time scales is the hyperspace $\mathrm{CL}(\mathbb{R})$. There are three popular topologies on hyperspaces: the induced topology by the Hausdorff metric, the Vietoris topology, and the Fell topology. We note that when interpreting a time scale $\mathbb{T}$ as a metric space we will not use the standard metric $d(x,y)=|x-y|$ but an equivalent bounded metric $d(x,y)=\min\{1,|x-y|\}$. It is known that the topology generated by this bounded metric is equivalent to the topology generated by the standard metric inherited from $\mathbb{R}$.

Which topology should be used on $\mathrm{CL}(\mathbb{R})$?

Example: Let us consider $\mathrm{CL}(\mathbb{R})$ with the topology induced by the Hausdorff metric. We will show that the time scales $[0,n]$ do not converge to $[0,\infty)$ as expected using the Hausdorff metric on $\mathrm{CL}(\mathbb{R})$. Let $n \in \mathbb{N}$. We can compute $$\begin{array}{ll} H_d([0,n],[0,\infty))&= \max \left\{ \sup_{a \in [0,n]} \inf_{b \in [0,\infty)} d(a,b), \sup_{b \in [0,\infty)}\inf_{a \in [0,n]} d(a,b) \right\} \\ &= \max \left\{ 0, 1 \right\} \\ &= 1. \end{array}$$ So we see that $$\displaystyle\lim_{n \rightarrow \infty} H_d([0,n],[0,\infty)) = \displaystyle\lim_{n \rightarrow \infty} 1 = 1,$$ implying that $[0,n]$ does not converge to $[0,\infty)$ in the topological space $(\mathrm{CL}(\mathbb{R}),\tau)$ where $\tau$ is the topology generated by the metric $H_d$.

Example: We will show that the time scales $\mathbb{T}_k = \left\{ n + \dfrac{1}{k} \colon n \in \mathbb{Z} \right\}$ does not converge to $\mathbb{Z}$ as $n \rightarrow \infty$ as expected under the Vietoris topology. Recall a sequence $x_n$ in a topological space converges to $x_0$ if for every open set $U$ containing $x_0$, there is some $N$ so that for all $n \geq N$, $x_n \in U$. Consider the set $$U = \displaystyle\bigcup_{k=1}^{\infty} \left( k - \dfrac{1}{k}, k + \dfrac{1}{k} \right),$$ which is a union of open intervals around the integers whose diameter converges to $0$ as $n \rightarrow \infty$. The set $U$ is an open set in $\mathbb{R}$. The set $U^+$ which is open in $(\mathrm{CL}(\mathbb{R}),\tau_v)$ (where $\tau_v$ denotes Vietoris toplogy) is given by the formula $$U^+ = \{A \in \mathrm{CL}(\mathbb{R}) \colon A \subset U\}.$$ Notice that $\mathbb{Z} \in U^+$. Let $n>1$, then $\mathbb{Z}+\dfrac{1}{n} \not\in U^+$ because for any $m>n$, $m+\dfrac{1}{n} > m+\dfrac{1}{m}$ and so $m+\dfrac{1}{n} \not\in \left( m - \dfrac{1}{m}, m+\dfrac{1}{m} \right)$. Therefore it is not possible for $\mathbb{Z}+\dfrac{1}{n}$ to converge to $\mathbb{Z}$ in the Vietoris topology.

References

@inproceedings{MR2547668,
  title="The Fell topology on the space of time scales for dynamic equations",
  author="Oberste-Vorth, Ralph W.",
  booktitle="Advances in Dynamical Systems and Applications 2008",
}

</bibtex>