Empty Set is Subset of All Sets/Proof 2

Theorem

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

That is:

$\forall S: \O \subseteq S$

Proof

$S \subseteq T$ means:

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

or, equivalently:

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

Thus:

\(\ds \)

\(\)

\(\ds S \subseteq T\)

\(\ds \)

\(\leadstoandfrom\)

\(\ds \forall x \in S: x \in T\)

Definition of Subset

\(\ds \)

\(\leadstoandfrom\)

\(\ds \neg \paren {\exists x \in S: \neg \paren {x \in T} }\)

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

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 3$: Unordered Pairs
• 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.1$: Sets and Subsets