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!


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 $B^c=B^c \cap X$

  • $\begingroup$ aarrghh.. thanks! missed that big time $\endgroup$ – nate Dec 17 '11 at 18:28

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