Difference of Sets

Difference of sets. The difference of two sets $A$ and $B$ (in that order), written

$A - B$ (also read "$A$ minus $B$"), is the set of elements that belong to $A$ but do

not belong to $B$:

Not commutative. Unlike union and intersection, set difference is generally

not commutative — that is, $A - B \ne B - A$ in general, because the two sets

describe entirely different things: $A - B$ keeps only $A$'s "private" elements, while

$B - A$ keeps only $B$'s "private" elements.

Worked illustration. Let $A = \{a, b, c, d, e\}$ and $B = \{b, d, f, g\}$. Then

Clearly $A - B \ne B - A$ here, confirming that difference is not commutative.

$A - B$ and $B - A$ are always disjoint. For any two sets $A$ and $B$,

Proof. Suppose, for contradiction, some element $x$ belonged to both $A - B$ and

$B - A$. Since $x \in A - B$, we have $x \in A$ and $x \notin B$. Since

$x \in B - A$, we have $x \in B$ and $x \notin A$. But "$x \in A$" and "$x \notin A$"

cannot both be true at once — a contradiction. Hence no such $x$ exists, so

$(A - B) \cap (B - A) = \varnothing$. $\blacksquare$ This matches the Venn-diagram

picture directly: $A - B$ is the part of circle $A$ outside the overlap, and $B - A$

is the part of circle $B$ outside the overlap — two regions that, by construction,

never touch.

Relation to a Venn diagram. In the two-circle Venn diagram of Section 1.5, $A - B$ …