3: A Classification of Points
- Page ID
- 204821
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For any element of any time scale the following classification holds.
For \(t \in \mathbb{T}\) we have the following cases.
1. If \(\sigma(t)>t\), then we say that \(t\) is right-scattered.
2. If \(t<\sup \mathbb{T}\) and \(\sigma(t)=t\), then we say that \(t\) is right-dense.
3. If \(\rho(t)<t\), then we say that \(t\) is left-scattered.
4. If \(t>\inf \mathbb{T}\) and \(\rho(t)=t\), then we say that \(t\) is left-dense.
5. If \(t\) is left-scattered and right-scattered at the same time, then we say that \(t\) is isolated.
6. If \(t\) is left-dense and right-dense at the same time, then we say that \(t\) is dense.
Let \(\mathbb{T}=h \mathbb{Z}, h>0\). By Example 0.10, we have that
\[\begin{aligned}
\sigma(t) & =t+h \\
& >t, \quad t \in \mathbb{T} .
\end{aligned}\notag \]
Thus, any point of \(\mathbb{T}\) is right-scattered. Now, using Example 0.30, we get
\[\begin{aligned}
\rho(t) & =t-h \\
& <t, \quad t \in \mathbb{T} .
\end{aligned}\notag \]
Therefore any point of \(\mathbb{T}\) is left-scattered. Hence, we conclude that any point of \(\mathbb{T}\) is isolated.
Let \(\mathbb{T}=3^{\mathbb{N}_0}\). Take \(t \in \mathbb{T}\) arbitrarily. We have the following cases.
1. Assume that \(t=1\). By Example 0.11 , we have
\[\begin{aligned}
\sigma(1) & =3 \\
& >1,
\end{aligned}\notag \]
i.e., \(t=1\) is right-scattered. By Example 0.31, we have
\[\sigma(1)=1 .\notag \]
Since \(1=\inf \mathbb{T}\), we conclude that \(t=1\) is not left-dense.
2. Let \(t>1\). By Example 0.11, we have
\[\sigma(t)=3 t\notag \]
\[>t .\notag \]
Thus, \(t\) is right-scattered. By Example 0.31, we get
\[\begin{aligned}
\rho(t) & =\dfrac{t}{3} \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left scattered. Hence, we conclude that \(t\) is isolated.
Let \(\mathbb{T}=\mathbb{N}_0^k, k \in \mathbb{N}\). Take \(t \in \mathbb{T}\) arbitrarily. We have the following cases.
1. Let \(t=0\). By Example 0.12, we have
\[\begin{aligned}
\sigma(0) & =1 \\
& >1,
\end{aligned}\notag \]
i.e., \(t=0\) is right-scattered. By Example 0.32, we obtain
\[\rho(0)=0 .\notag \]
Since \(0=\inf \mathbb{T}\), we conclude that \(t=0\) is not left-dense.
2. Let \(t>0\). By Example 0.12, we get
\[\begin{aligned}
\sigma(t) & =(\sqrt[k]{t}+1)^k \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. B y Example 0.32, we find
\[\begin{aligned}
\rho(t) & =(\sqrt[k]{t}-1)^k \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Therefore \(t\) is isolated.
Let \(\mathbb{T}=\left\{H_n: n \in \mathbb{N}_0\right\}\), where \(H_n, n \in \mathbb{N}_0\), are the harmonic numbers. Take \(n \in \mathbb{N}_0\). We have the following cases.
1. Let \(n=0\). Then, by Example 0.13 , we get
\[\sigma\left(H_0\right)=H_1,\notag \]
i.e., \(H_0\) is right-scattered. By Example 0.33, we have
\[\rho\left(H_0\right)=H_0 .\notag \]
Since \(H_0=\inf \mathbb{T}\), we conclude that \(H_0\) is not left-dense.
2. Let \(n>0\). By Example 0.13, we get
\[\begin{aligned}
\sigma\left(H_n\right) & =H_{n+1} \\
& >H_n .
\end{aligned}\notag \]
Then \(H_n\) is right-scattered. By Example 0.33, we get
\[\begin{aligned}
\rho\left(H_n\right) & =H_{n-1} \\
& <H_n
\end{aligned}\notag \]
i.e., \(H_n\) is left-scattered. Then \(H_n\) is isolated.
Let \(\mathbb{T}=P_{1,3}\). Take \(t \in \mathbb{T}\) arbitrarily. We have the following cases.
1. Let \(t \in \bigcup_{k=0}^{\infty}\). By Example 0.14 , we get
\[\sigma(t)=t,\notag \]
i.e., \(t\) is right-dense. By Example 0.34, we find
\[\rho(t)=t,\notag \]
i.e., \(t\) is left-dense. Thus, \(t\) is dense.
2. Let \(t=0\). By Example 0.14 , we obtain
\[\sigma(0)=0,\notag \]
i.e, \(t=0\) is right-dense. By Example 0.34, we find
\[\rho(0)=0 .\notag \]
Since \(0=\inf \mathbb{T}\), we conclude that 0 is not left-dense.
3. Let \(t \in \bigcup_{k=1}\{4 k\}\). By Example 0.14, we get
\[\sigma(t)=t,\notag \]
i.e., \(t\) is right-dense. By Example 0.34, we find
\[\rho(t)=t-3\notag \]
\(<t\),
i.e., \(t\) is left-scattered.
4. Let \(t \in \bigcup_{k=0}^{\infty}\{4 k+1\}\). By Example 0.14, we find
\[\begin{aligned}
\sigma(t) & =t+3 \\
& >t,
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.34, we find
\[\rho(t)=t\notag \]
i.e., \(t\) is left-dense.
Let \(\mathbb{T}=C\), where \(C\) is the Cantor set. Take \(t \in \mathbb{T}\) arbitrarily. We have the following cases.
1. Let \(t \in C_1\). By Example 0.15, we have
\[\begin{aligned}
\sigma(t) & =t+\dfrac{1}{3^{m+1}} \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.35, we find
\[\rho(t)=t .\notag \]
If \(t \neq 0\), then it is left-dense. If \(t=0\), then it is not left-dense because \(0=\inf \mathbb{T}\).
v2 . Let \(t \in C_2\). By Example 0.15, we get
\[\sigma(t)=t,\notag \]
i.e., \(t\) is right-dense. By Example 0.35, we find
\[\begin{aligned}
\rho(t) & =t-\dfrac{1}{3^{m+1}} \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered.
3. Let \(t \in T \backslash C_1\). We have the following subcases.
a. Let \(t \in C_2\). By Example 0.15, we find
\[\sigma(t)=t\notag \]
i.e., \(t\) is right-dense. By Example 0.35, we obtain
\[\rho(t)=t-\dfrac{1}{3^{m+1}}\notag \]
\[<t,\notag \]
i.e., \(t\) is left-scattered.
b. Let \(t \in T \backslash C_2\). By Example 0.15, we arrive at
\[\sigma(t)=t,\notag \]
i.e., \(t\) is right-dense. By Example 0.35, we have
\[\rho(t)=t .\notag \]
If \(t \neq 0\), then it is left-dense. If \(t=0\), then it is not left-dense.
4. Let \(t \in \mathbb{T} \backslash C_2\). We have the following subcases.
a. Let \(t \in C_1\). By Example 0.15 , we find
\[\begin{aligned}
\sigma(t) & =t+\dfrac{1}{3^{m+1}} \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.35, we find
\[\rho(t)=t .\notag \]
If \(t \neq 0\), then it is left-dense. If \(t=0\), then is left-dense.
b. Let \(t \in T \backslash C_1\). By Example 0.15, we have
\[\sigma(t)=t,\notag \]
i.e., \(t\) is right-dense. By Example 0.35, we have
\[\rho(t)=t .\notag \]
If \(t \neq 0\), then it is left-dense and hence dense. If \(t=0\), it is not left-dense
Let
\[\mathbb{T}=\left\{\sum_{k=0}^{n-1} \alpha_k: \alpha_k>0, \quad k \in \mathbb{N}_0, \quad n \in \mathbb{N}\right\}\notag \]
Take \(t \in \mathbb{T}\) arbitrarily. Then there is a \(n \in \mathbb{N}\) such that
\[t=\sum_{k=0}^{n-1} \alpha_k\notag \]
We have the following cases.
1. Let \(n=1\). By Example 0.16, we have
\[\begin{aligned}
\sigma(t) & =\sum_{k=0}^n \alpha_k \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.36, we arrive at
\[\rho(t)=t .\notag \]
Since \(t=\inf ^{\top} \mathbb{T}\), we conclude that \(t\) is not left-dense.
2 . Let \(n>1\). By Example 0.16, we find
\[\begin{aligned}
\sigma(t) & =\sum_{k=0}^n \alpha_k \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.36, we find
\[\begin{aligned}
\rho(t) & =t-\alpha_{n-1} \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Therefore \(t\) is isolated.
Let \(\mathbb{T}=\left\{t_n=-\dfrac{1}{n}: n \in \mathbb{N}\right\} \cup \mathbb{N}_0\). Take \(t \in \mathbb{T}\) arbitrarily. We have the following cases.
1. Let \(t=-1\). By Example 0.17, we have
\[\begin{aligned}
\sigma(-1) & =-\dfrac{1}{2} \\
& >-1
\end{aligned}\notag \]
i.e., \(t=-1\) is right-scattered. By Example 0.37, we find
\[\rho(-1)=-1 .\notag \]
Since \(-1=\inf ^T \mathbb{T}\), we conclude that \(t=-1\) is not left-dense.
2 . Let \(t=0\). By Example 0.17, we have
\[\begin{aligned}
\sigma(0) & =1 \\
& >0
\end{aligned}\notag \]
i.e., \(t=0\) is right-scattered. By Example 0.37, we obtain
\[\rho(0)=0 .\notag \]
Since \(0>\inf \mathbb{T}\), we conclude that \(t=0\) is left-dense.
3. Let \(t \in\left\{t_n=-\dfrac{1}{n}: n \in \mathbb{N}\right\} \backslash\{-1\}\). By Example 0.17 , we have
\[\begin{aligned}
\sigma(t) & =-\dfrac{t}{t-1} \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.37, we get
\[\begin{aligned}
\rho(t) & =\dfrac{t}{t+1} \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
4. Let \(t \in \mathbb{N}\). By Example 0.17, we have
\[\begin{aligned}
\sigma(t) & =t+1 \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.37, we obtain
\[\begin{aligned}
\rho(t) & =t-1 \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
Let \(\mathbb{T}=\left\{t_n=\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}_0\right\} \bigcup\{0,1\}\). Take \(t \in \mathbb{T}\) arbitrarily. We have the following cases.
1. Let \(t=0\). By Example 0.18, we have
\[\sigma(0)=0\notag \]
i.e., \(t=0\) is right-dense. By Example 0.38, we get
\[\rho(0)=0\notag \]
Since \(0=\inf \mathbb{T}\), we conclude that \(t=0\) is not left-dense.
2 . Let \(t=\dfrac{1}{2}\). By Example 0.18 , we have
\[\sigma\left(\dfrac{1}{2}\right)=1\notag \]
i.e., \(t=\dfrac{1}{2}\) is right-scattered. By Example 0.38, we get
\[\begin{aligned}
\rho\left(\dfrac{1}{2}\right) & =\dfrac{1}{4} \\
& <\dfrac{1}{2}
\end{aligned}\notag \]
i.e., \(t=\dfrac{1}{2}\) is left-scattered. Thus, \(t=\dfrac{1}{2}\) is isolated.
3. Let \(t=1\). By Example 0.18, we have
\[\sigma(1)=1 .\notag \]
Since \(1=\sup \mathbb{T}\), we conclude that \(t=1\) is not right-dense. By Example 0.38, we obtain
\[\begin{aligned}
\rho(1) & =\dfrac{1}{2} \\
& <1
\end{aligned}\notag \]
i.e., \(t=1\) is left-scattered.
4. Let \(t \in\left\{t_n=\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}\right\}\). Then there is a \(n \in \mathbb{N}\) such that
\[t=\left(\dfrac{1}{2}\right)^{2^n}\notag \]
By Example 0.18, we have
\[\begin{aligned}
\sigma(t) & =\sqrt{t} \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.38, we obtain
\[\begin{aligned}
\rho(t) & =t^2 \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
Let \(U=\left\{\dfrac{1}{2^n}: n \in \mathbb{N}\right\}\) and
\[\mathbb{T}=\{0\} \cup U \cup(1-U) \cup(1+U) \cup(2-U) \cup(2+U) \cup\{1,2\} .\notag \]
Take \(t \in \mathbb{T}\) arbitrarily. Then, we have the following cases.
1. Let \(t=0\). By Example 0.19 , we have
\[\sigma(0)=0\notag \]
i.e., \(t=0\) is right-dense, By Example 0.39, we get
\[\rho(0)=0\notag \]
Since \(0=\inf \mathbb{T}\), we conclude that \(t=0\) is not left-dense.
2 . Let \(t=\dfrac{1}{2}\). By Example 0.19 , we have
\[\sigma\left(\dfrac{1}{2}\right)=\dfrac{3}{4},\notag \]
i.e., \(t=\dfrac{1}{2}\) is right-scattered. By Example 0.39, we get
\[\rho\left(\dfrac{1}{2}\right)=\dfrac{1}{4}\notag \]
i.e., \(t=\dfrac{1}{2}\) is right-scattered. Thus, \(t=\dfrac{1}{2}\) is isolated.
3. Let \(t=1\). By Example 0.19, we have
\[\sigma(1)=1,\notag \]
i.e., \(t=1\) is right-dense. By Example 0.39, we find
\[\rho(1)=1,\notag \]
i.e., \(t=1\) is left-dense. Thus, \(t=1\) is dense.
4 . Let \(t=\dfrac{3}{2}\). By Example 0.19 , we have
\[\sigma\left(\dfrac{3}{2}\right)=\dfrac{7}{4}\notag \]
i.e., \(t=\dfrac{3}{2}\) is right-scattered. By Example 0.39, we find
\[\rho\left(\dfrac{3}{2}\right)=\dfrac{5}{4}\notag \]
i.e., \(t=\dfrac{3}{2}\) is left-scattered. Thus, \(t=\dfrac{3}{2}\) is isolated.
5 . Let \(t=2\). By Example 0.19, we have
\[\sigma(2)=2,\notag \]
i.e., \(t=2\) is right-dense. By Example 0.39, we get
\[\rho(2)=2\notag \]
i.e., \(t=2\) is left-dense. Thus, \(t=2\) is dense.
6. Let \(t=\dfrac{5}{2}\). Then, by Example 0.19 , we get
\[\sigma\left(\dfrac{5}{2}\right)=\dfrac{5}{2}\notag \]
Since \(\dfrac{5}{2}=\sup \mathbb{T}\), we conclude that \(t=\dfrac{5}{2}\) is not right-dense. By Example 0.39 , we find
\[\rho\left(\dfrac{5}{2}\right)=\dfrac{9}{8}\notag \]
Thus, \(t=\dfrac{5}{2}\) is left-scattered.
7. Let \(t \in U \backslash\left\{\dfrac{1}{2}\right\}\). By Example 0.19 , we have
\[\begin{aligned}
\sigma(t) & =2 t \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.39, we get
\[\begin{aligned}
\rho(t) & =\dfrac{t}{2} \\
& <t,
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
8. Let \(t \in(1-U) \backslash\left\{\dfrac{1}{2}\right\}\). By Example 0.19, we have
\[\begin{aligned}
\sigma(t) & =\dfrac{1+t}{2} \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.39, we find
\[\begin{aligned}
\rho(t) & =2 t-1 \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
9. Let \(t \in(1+U) \backslash\left\{\dfrac{3}{2}\right\}\). By Example 0.19 , we have
\[\sigma(t)=2 t-1\notag \]
\[>t\notag \]
i.e., \(t\) is right-scattered. By Example 0.39, we find
\[\begin{aligned}
\rho(t) & =\dfrac{t+1}{2} \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
10. Let \(t \in(2-U) \backslash\left\{\dfrac{3}{2}\right\}\). By Example 0.19, we have
\[\begin{aligned}
\sigma(t) & =\dfrac{2+t}{2} \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.39, we find
\[\begin{aligned}
\rho(t) & =2 t-2 \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
11. Let \(t \in(2+U) \backslash\left\{\dfrac{5}{2}\right\}\). By Example 0.19 , we have
\[\begin{aligned}
\sigma(t) & =2(t-1) \\
& >t
\end{aligned}\notag \]
i.e., \(t\) is right-scattered. By Example 0.39, we find
\[\begin{aligned}
\rho(t) & =\dfrac{t+2}{2} \\
& <t
\end{aligned}\notag \]
i.e., \(t\) is left-scattered. Thus, \(t\) is isolated.
Classify each point of \(\mathbb{T}\), where \(\mathbb{T}\) are the time scales in Exercise 0.3.