Cardinalities form Inequality implies Difference is Nonempty

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Theorem

Let $X, Y$ be sets.

Let

$\card X < \card Y$

where $\card X$ denotes the cardinality of $X$.

Then:

$Y \setminus X \ne \O$


Proof

Aiming for a contradiction, suppose that:

$Y \setminus X = \O$

Then by Set Difference with Superset is Empty Set:

$Y \subseteq X$

Hence by Subset implies Cardinal Inequality:

$\card Y \le \card X$

This contradicts:

$\card X < \card Y$

Hence the result.

$\blacksquare$


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