Intersection with Complement

Theorem

The intersection of a set and its complement is the empty set:

$S \cap \map \complement S = \O$

Proof

Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cap \relcomp S T = \O$ from Intersection with Relative Complement is Empty.

$\blacksquare$

Also see

• Union with Complement
• Symmetric Difference with Complement

Notice the similarity with the Principle of Non-Contradiction.

The complement of a set is similar to the negation of a proposition, and intersection is similar to conjunction.

This article, or a section of it, needs explaining. In particular: The above comment lacks precision. There is a direct relationship between these results which can be established in the context of abstract algebra, which is the preferred approach. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{B v}$
• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(i)}$
• 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): $\S 2$
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): algebra of sets: $\text {(vii)}$
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): algebra of sets: $\text {(vii)}$