Set Union is not Cancellable

Theorem

Set union is not a cancellable operation.

That is, for a given $A, B, C \subseteq S$ for some $S$, it is not always the case that:

$A \cup B = A \cup C \implies B = C$

Proof

Proof by Counterexample:

Let $S = \set {a, b}$.

Let:

$A = \set {a, b}$

$B = \set a$

$C = \set b$

Then:

\(\ds A \cup B\)

\(=\)

\(\ds \set {a, b}\)

\(\ds \)

\(=\)

\(\ds A \cup C\)

but:

\(\ds B\)

\(=\)

\(\ds \set a\)

\(\ds \)

\(\ne\)

\(\ds \set b\)

\(\ds \)

\(=\)

\(\ds C\)

$\blacksquare$

Sources

• 1964: William K. Smith: Limits and Continuity ... (previous) ... (next): $\S 2.1$: Sets: Exercise $\text{B} \ 2$