Empty Set is Unique
Theorem
Proof
Let $\O$ and $\O'$ both be empty sets.
From Empty Set is Subset of All Sets, $\O \subseteq \O'$, because $\O$ is empty.
Likewise, we have $\O' \subseteq \O$, since $\O'$ is empty.
Together, by the definition of set equality, this implies that $\O = \O'$.
Thus there is only one empty set.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 3$: Unordered Pairs
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.1: \ 1$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.3$: Subsets
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 6.5$: Subsets
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1.2$: Operations on Sets: Exercise $1.2.4$
- 1999: András Hajnal and Peter Hamburger: Set Theory ... (previous) ... (next): $1$. Notation, Conventions: $4$
