Set Intersection is not Cancellable

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Theorem

Set intersection 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:

$A \cap B = A \cap C \implies B = C$


Proof

Proof by Counterexample:

Let $S = \left\{ {a, b, c}\right\}$.

Let:

$A = \left\{ {a}\right\}$
$B = \left\{ {a, b}\right\}$
$C = \left\{ {a, c}\right\}$

Then:

\(\displaystyle A \cap B\) \(=\) \(\displaystyle \left\{ {a}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle A \cap C\)

but:

\(\displaystyle B\) \(=\) \(\displaystyle \left\{ {a, b}\right\}\)
\(\displaystyle \) \(\ne\) \(\displaystyle \left\{ {a, c}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle C\)

$\blacksquare$


Sources