Difference between revisions of "Isolated points"
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Revision as of 05:05, 18 May 2014
Let $X \subset \mathbb{R}$. We say a point $x \in X$ is an isolated point if there exists a $\delta > 0$ such that $(t-\delta,t+\delta) \cap X = \emptyset$. It is known that for any such $X$, the set of isolated points of $X$ is at most countable. Moreover a set of isolated points is closed in $\mathbb{R}$ because its complement is a union of open intervals.
Let $\mathbb{T}=\{\ldots,t_{-1},t_0,t_1,\ldots\}$ be a time scale of isolated points with $t_k > t_n$ iff $k>n$. Define the bijection $\pi \colon \mathbb{T} \rightarrow \mathbb{Z}$, $\pi(t_k)=k$.
The set $h\mathbb{Z}=\{\ldots,-2h,-h,0,h,2h,\ldots\}$ of multiples of the integers is a time scale.
| Generic element $t\in \mathbb{T}$: | For some $n \in \mathbb{Z}, t=t_n$ |
| Jump operator: | $\sigma(t)=\sigma(t_n)=t_{n+1}$ |
| Graininess operator: | $\mu(t)=\mu(t_n)=t_{n+1}-t_n$ |
| $\Delta$-derivative: | $f^{\Delta}(t)=f^{\Delta}(t_n) = \dfrac{f(t_{n+1})-f(t_n)}{t_{n+1}-t_n}$ |
| $\Delta$-integral: | $\displaystyle\int_s^t f(\tau) \Delta \tau = \displaystyle\sum_{k=s}^{t-1} f(k)$ |
| Exponential function: | $\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_{s}^{t} \dfrac{1}{\mu(\tau)} \log(1 + p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=s}^{t-1} \log(1+p(k)) \right) \\ &= \displaystyle\prod_{k=s}^{t-1} \left( 1+p(k) \right) \\ \end{array}$ |
