Halmos: Naive Set Theory, Powers and Complements

In Halmos's book under section 5, one of the definitions call for the proof of:

$$A \cap (B-C) = (A \cap B) - (A \cap C) = (A \cap B) \cap (A \cap C)^c$$

So I figured, ok, under the distributive property, one would obtain the following. Exploring the LHS I realize that I don't exactly get the equality, for example doing the approach through associativity:

$$A \cap (B \cap C^c) = (A \cap B) \cap C^c = (A \cap B) - C$$

that doesn't seem right, or following through the distributive property + De Morgans law:

$$A \cap (B \cap C^c) = (A \cap B) \cap (A \cap C^c)$$ $$(A \cap B) \cap (A \cap C^c) = (A \cap B)\cap (A^c \cup C)^c$$ $$(A \cap B)\cap (A^c \cup C)^c = (A \cap B)-(A^c \cup C)$$

Unless I'm missing some fact that involves $A \cap C \subset C$, or details regarding $x \in (A \cap C^c)$, I don't get this particular portion:

$$(A \cap B) - (A \cap C) \neq (A \cap B)-(A^c \cup C) \neq (A \cap B) - C$$

I do hope you can enlighten me, and confirm that I am indeed a damned fool (im pretty sure I missed something). Thanks.

P.S I also apologize in advance if there are similar questions.

For the left part, we have to use the definition of (relative) complement to get:

$A \cap (B \setminus C)=A \cap (B \cap C^c)$

and

$(A \cap B) \setminus (A \cap C)= (A \cap B) \cap (A \cap C)^c=(A \cap B) \cap (A^c \cup C^c)=[(A \cap B) \cap A^c] \cup [(A \cap B) \cap C^c]$.

But: $(A \cap B) \cap A^c = (A \cap A^c) \cap B = \emptyset \cap B = \emptyset$, and thus from the above we get:

$(A \cap B) \setminus (A \cap C)= \emptyset \cup [(A \cap B) \cap C^c] = (A \cap B) \cap C^c.$

It seems to me that your mistake is assuming $\ne$ in the last "equation":

$x \in (A \cap B) \setminus (A \cap C) \text { iff } (x \in (A \cap B) \text { and } x \notin (A \cap C)).$

Thus:

$\text { iff } (x \in A \text { and } x \in B) \text { and } ( \text { either } x \notin A \text { or } x \notin C).$

But the first alternative is ruled out, because we have already $x \in A$, and thus we have:

$\text {iff } (x \in A \text { and } x \in B) \text { and } x \notin C \text { i.e. } x \in (A \cap B) \setminus C$.

In the same vain:

$x \in (A \cap B) \setminus (A^c \cup C) \text { iff } ((x \in A \text { and } x \in B) \text { and } x \notin (A^c \cup C)) \text { iff } ((x \in A \text { and } x \in B) \text { and } (x \notin A^c \text{ and } x \notin C)).$

But clearly, if $x \in A$ then $x \notin A^c$, and thus we simplify to:

$\text { iff } (x \in A \text { and } x \in B) \text { and } x \notin C.$

• AH, right, so its the interpretation of my results at the end. Thank you very much. +1 – Cyrus Nov 22 '17 at 10:22