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A mapping $\Lambda:\mathbb{R}^{n+1}\rightarrow \mathbb{R}^m$ is said to be upper semi-continuous with respect to inclusion(u.s.c.i) at $\left(t_0,x_0\right)$, where, $t_0\in \mathbb{R}$ and $x_0\in\mathbb{R}$ if for every $\epsilon>0$, there exists $\delta > 0$ such that \begin{equation} \Lambda\left(t,x\right)\subseteq \left[\Lambda\left(t_0,x_0\right)\right]_{\epsilon} \end{equation} for every $\left(t,x\right)$ in $\delta-$ neighborhood of $\left(t_0,x_0\right)$. and $[U]_{\epsilon}$ represents closed $\epsilon$ neighborhood of $U$.

My doubt comes in this second part mentioned below:

"The mapping is said to be u.s.c.i. on $\mathbb{R}^{n+1}$ if it is u.s.c.i. at every point of $\mathbb{R}^{n+1}$".

Since the mapping of all points ( and these points may be in $\mathbb{R}^{n+1}$) in the $\delta-$ neighborhood of $\left(t_0,x_0\right)$ lie inside the $\epsilon-$ neighborhood of mapping of $\left(t_0,x_0\right)$, then clearly, for all these points inside $B\left((t_0,x_0),\delta\right)$, barring $(t_0,x_0)$, the requirement for u.s.c.i. is not satisfied, unless the $\subseteq$ sign is replaced with $=$. Otherwise, it is not possible that $\Lambda$ will be u.s.c.i. on every part of $\mathbb{R}^{n+1}$.

Can someone explain in detail, the meaning of sentence written in quotes.

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  • $\begingroup$ It seems $\Lambda$ is perhaps meant to be $\mathcal P(\mathbb R^{n+1}) \to \mathcal P(\mathbb R^m)$? $\endgroup$ – Anthony Carapetis Nov 5 '14 at 2:34
  • $\begingroup$ Here is the definition of u.s.c.i. by the author: books.google.com/… $\endgroup$ – user146290 Nov 5 '14 at 2:37
  • $\begingroup$ So it seems $\Lambda : \mathbb R^{n+1} \to \mathcal P (\mathbb R^m)$. $\endgroup$ – Anthony Carapetis Nov 5 '14 at 3:00
  • $\begingroup$ I am new to analysis. What does $\mathcal{P}$ denote? $\endgroup$ – user146290 Nov 5 '14 at 3:06
  • $\begingroup$ The power set. In words, $\Lambda$ maps elements of $\mathbb R^{n+1}$ to subsets of $\mathbb R^m$. $\endgroup$ – Anthony Carapetis Nov 5 '14 at 4:08

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