# Empty Set is Subset of All Sets/Proof 2

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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

$S \subseteq T$ 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