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 $\mathfrak{M}$)

  • $\begingroup$ 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

$P(X)$ 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".

  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.