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