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:

\(\displaystyle S \subseteq T\) \(\iff\) \(\displaystyle \forall x \in S: x \in T\) $\quad$ Definition of Subset $\quad$
\(\displaystyle \) \(\iff\) \(\displaystyle \neg \left({\exists x \in S: \neg \left({x \in T}\right)}\right)\) $\quad$ De Morgan's Laws (Predicate Logic) $\quad$

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


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

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

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

$\blacksquare$


Sources