Empty Set is Subset of All Sets

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Theorem

The empty set $\O$ is a subset of every set (including itself).

That is:

$\forall S: \O \subseteq S$


Proof 1

By the definition of subset, $\O \subseteq S$ means:

$\forall x: \paren {x \in \O \implies x \in S}$

By the definition of the empty set:

$\forall x: \neg \paren {x \in \O}$

Thus $\O \subseteq S$ is vacuously true.

$\blacksquare$


Proof 2

$S \subseteq T$ means:

every element of $S$ is also in $T$

or, equivalently:

every element that is not in $T$ is not in $S$ either.


Thus:

\(\displaystyle \) \(\) \(\displaystyle S \subseteq T\)
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \forall x \in S: x \in T\) Definition of Subset
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \neg \paren {\exists x \in S: \neg \paren {x \in T} }\) De Morgan's Laws (Predicate Logic)

which means there is no element in $S$ which is not also in $T$.


There are no elements of $\O$, from the definition of the empty set.

Therefore $\O$ has no elements that are not also in any other set.

Thus, from the above, all elements of $\O$ are all (vacuously) in every other set.

$\blacksquare$


Sources