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Let $\mu (V) = V_1 \cup V_2 \cup \cdots \cup V_k$, where each $|V_i| \le n$; means each $V_i$ contains at most $n$ points and Intersection between any two $V_i$ and $V_j$ is empty.

What does $P(\mu (V))$ mean? Does it mean $P(V_1,V_2, \ldots, V_k)$ or anything else? Is it true that a typical element in power-set will be $\{V_1,V_2\}$. Are these two valid elements in a power -set? enter image description here

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  • $\begingroup$ No. Elements of the power set will contain members of the $V_n$'s, not the $V_n$'s themselves. $\endgroup$ – Malice Vidrine Sep 17 '17 at 17:00
  • $\begingroup$ I wonder if you saw $\bigcup V$ and copied it as $\mu V. \qquad$ $\endgroup$ – Michael Hardy Sep 17 '17 at 17:22
  • $\begingroup$ @ Michael Hardy No $\endgroup$ – user437890 Sep 17 '17 at 17:51
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By what you wrote, $\mu(V)$ consists of all elements which appear in one of the $V_i$'s. Consequently, elements of the powerset $P(\mu(V))$ are just sets $U$ such that all elements of $U$ are elements of $V_i$'s. In particular, $\{V_1,V_2\}$ is (usually) not be an element of the powerset, but for example $V_1\cup V_2$ is.

EDIT: Since you added a new question with a picture in your original post: Yes, these two sets drawn there are elements of the powerset, since they contain only elements which are also elements of the $V_i$'s. The second set drawn there is $V_4\cup V_5$, which is close to the example I gave.

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    $\begingroup$ Does all the elements of particular layer has to be there or some of them is also allowed? $\endgroup$ – user437890 Sep 17 '17 at 17:11
  • $\begingroup$ This is allowed, for example the set 1 in your picture. It has to by any collection of elements of the $V_i$'s. $\endgroup$ – x432ph Sep 17 '17 at 17:14

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