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Let $X$ be a set and $\mathfrak{M}\subset \mathcal{P}(x)$ be a family of sets over(?) $X$.

Does this simply mean, that if for example $X$ contains all squares of a chess board, any set $A\subset\mathfrak{M}$ is an arbitary combination of squares from this board?

And $\mathfrak{M}$ are all possible subsets of $X$?

Edit: oh, i mean does $\mathcal{P}(x)$ contain any possible subset of X? (instead of $\mathfrak{M}$)

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  • $\begingroup$ A family of sets over $X$ is a set of subsets of $X$ (recall that $\mathcal P(X)$ is the power-set of $X$, i.e. the set of all subsets of $X$. $\endgroup$ – Mauro ALLEGRANZA Jul 8 '16 at 13:05
  • $\begingroup$ IF $\subset$ means "strictly included", this means that $M \ne \mathcal P(X)$. $\endgroup$ – Mauro ALLEGRANZA Jul 8 '16 at 13:06
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$P(X)$ denotes the power set of $X$, i.e. the collection of all possible subsets of $X$:

\begin{equation} P(X) := \{A : A \subseteq X \} \end{equation}

Then a familiy of sets $\mathfrak{M} \subset P(X)$ is simply a set that contains other sets, and in our case the contained sets in $\mathfrak{M}$ are subsets of $X$. "Family" can be seen as another word for "set", so one doesn't have to say $\mathfrak{M}$ is a "set of sets".

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  • $\begingroup$ so if i have the set $X=\{1,2\}$, then $P(X)=\{\emptyset ,\{1\},\{1,2\},\{2\}\}$ right? or do we not include the empty set in the powerset? $\endgroup$ – SAJW Jul 8 '16 at 13:18
  • $\begingroup$ @saturatedexpo Yes you're right, the empty set is always a subset of a set $X$ and thus contained in the power set. If you want to check if you have the right number of sets in $P(X)$ when building it, you can use that $|P(X)| = 2^{|X|}$. $\endgroup$ – user331406 Jul 8 '16 at 13:21
  • $\begingroup$ do you refer in your equation to the cardinality? wow, that's a nice formular then :) (so X with 9 elements would have a Powerset with 81 elements? ) $\endgroup$ – SAJW Jul 8 '16 at 13:23
  • $\begingroup$ @saturatedexpo Yes, this is what I mean. Another notation for cardinality is sometimes $\#X$. The formula has a simple proof, because if you want to build a subset $A \subseteq X$, you can choose for every element $x \in X$ if you want to have it in $A$ or not, so you have $2^{|X|}$ possibilites to build your set $A$, and hence $|P(X)| = 2^{|X|}$. Note that this formula only holds if your set $X$ is finite. $\endgroup$ – user331406 Jul 8 '16 at 13:28
  • $\begingroup$ oh, i misread your equation, i thought $|x|^2$, just a "mind-typo" from me. thanks for the explanation. $\endgroup$ – SAJW Jul 8 '16 at 13:32
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This just means that $\mathfrak M$ is a collection of subsets of $X$.

The smallest collection of subsets of $X$ is $$\varnothing = \{\}$$ The largest collection of subsets of $X$ is $$\mathcal{P}(X)=\{A:A\subset X\}$$ So $\mathfrak M$ is somewhere in between these (and could possibly be one of them).

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