Distributing Over Sets

Relevant wiki: Sets - Multiple sets

For example, let $A=\{ a\} \\ B=\{ b\} \\ C=\{ c\}$

The union $A\cup(B-C)$ is the set $\{ a, b\}$ .

The set $A\cup B=\{ a,b\}$ and the set $A\cup C=\{ a,c\}$ .

So their set difference is $\{ a,b\} -\{ a,c\} =\{ b\}$ , which is not equal to the set $\{ a, b\}$ .

In general, for cases where $A$ is nonempty, the union $A\cup(B-C)$ will be distinct from the set difference $(A\cup B)-(A\cup C)$ because all members of set $A$ are in the former, and no elements of $A$ are in the latter.

Note: Here, the notation $X - Y$ refers to the relative complement of the set $X$ with the set $Y$ , also notated as $X \setminus Y$ .

https://brilliant.org/discussions/thread/venn-diagrams-and-set-notation/ There exist three nonempty sets A, B, and C such that A∪(B−C) ≠(A∪B)−(A∪C). When {a,b,d,g,e} ≠ {b}
From the diagram and by definitions:

A={a,d,e,g}, B={b,d,f,g}, C={c,e,f,g}

(B−C)={b,d,f,g}-{c,e,f,g}={b,d}

(A∪B)={a,d,e,g}∪{b,d,f,g}={a,b,d,e,f,g}

(A∪C)={a,d,e,g}∪{c,e,f,g}={a,c,d,e,f,g}

A∪(B−C)= {a,d,g,e} ∪ {b,d}= {a,b,d,g,e}

(A∪B)−(A∪C) = {a,b,d,e,f,g}-{a,c,d,e,f,g}= {b}