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I want to show that the following provisions satisfy an algebra (not a sigma algebra):

(a) $X$ is the non-empty set and $\mathcal{A}$ is the collection of subsets.

(b) $X \in \mathcal{A}$

(c) $A,B \in \mathcal{A}$ implies that $A \cap B^c \in \mathcal{A}$


So far I have:

If $A,B \in \mathcal{A}$, then so is their union and intersection, which means then that their symmetric difference is too,

$\ast$ $A \cup B - A \cap B = A \triangle B (= A \backslash B \cup B \backslash A)$.

I also know that, because of the implication in (c), $A \cap B^c = A-B = A \backslash B = (A^c \cup B)^c$, which takes care of all terms on left hand side, and the first term on the right hand side of $\ast$. So I think this means that both $A-B$ and $B-A$ are in the collection.

Does this mean $A^c$ and $B^c$ are too?

I mean $A-B$ is the complement of $B$ intersected with everything else $(A-B = B^c \cap else)$ so I am wondering if I need to use some relative set theory algebra with more sets, like $C$ and $D$?

I know that the requirements of an algebra are 1.) $X \in \mathcal{A}$ 2.) closed under complementation 3.) closed under finite union 4.) closed under finite intersection and that the last two are related via the second one.

Thanks for any suggestions!

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1 Answer 1

up vote 1 down vote accepted

The property of algebra are the first 3 that you wrote (you can derive the 4th by the first 3 and De Morgan).

If $B \in \mathcal{A}$ then $B^c \in \mathcal{A}$ because in your c) take $B=B$ and $A=X$, so $B^c=B^c \cap X$

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aarrghh.. thanks! missed that big time –  nate Dec 17 '11 at 18:28

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