Set Complement inverts Subsets
Theorem
Let $S$ and $T$ be sets.
Then:
- $S \subseteq T \iff \map \complement T \subseteq \map \complement S$
where:
- $S \subseteq T$ denotes that $S$ is a subset of $T$
- $\complement$ denotes set complement.
Corollary
- $S \subseteq \map \complement T \iff T \subseteq \map \complement S$
Proof 1
| \(\ds S\) | \(\subseteq\) | \(\ds T\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds S \cap T\) | \(=\) | \(\ds S\) | Intersection with Subset is Subset | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement {S \cap T}\) | \(=\) | \(\ds \map \complement S\) | Complement of Complement | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement S \cup \map \complement T\) | \(=\) | \(\ds \map \complement S\) | De Morgan's Laws: Complement of Intersection | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement T\) | \(\subseteq\) | \(\ds \map \complement S\) | Union with Superset is Superset |
$\blacksquare$
Proof 2
| \(\ds S\) | \(\subseteq\) | \(\ds T\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds (x \in S\) | \(\implies\) | \(\ds x \in T)\) | Definition of Subset | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds (x \notin T\) | \(\implies\) | \(\ds x \notin S)\) | Rule of Transposition | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds (x \in \map \complement T\) | \(\implies\) | \(\ds x \in \map \complement S)\) | Definition of Set Complement | ||||||||||
| \(\ds \map \complement T\) | \(\subseteq\) | \(\ds \map \complement S\) | Definition of Subset |
$\blacksquare$
Proof 3
By definition of set complement:
- $\map \complement T := \relcomp {\mathbb U} T$
where:
- $\mathbb U$ is the universal set
- $\relcomp {\mathbb U} T$ denotes the complement of $T$ relative to $\mathbb U$.
Thus the statement can be expressed as:
- $S \subseteq T \iff \relcomp {\mathbb U} T \subseteq \relcomp {\mathbb U} S$
The result then follows from Relative Complement inverts Subsets.
$\blacksquare$
Proof 4
| \(\ds S\) | \(\subseteq\) | \(\ds T\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds S\) | \(=\) | \(\ds S \cup T\) | Union with Superset is Superset | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement S\) | \(=\) | \(\ds \map \complement {S \cup T}\) | Complement of Complement | ||||||||||
| \(\ds \) | \(=\) | \(\ds \map \complement S \cap \map \complement T\) | De Morgan's Laws: Complement of Union | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map \complement S\) | \(\subseteq\) | \(\ds \map \complement T\) | Intersection with Subset is Subset |
$\blacksquare$
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{B x}$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(b)}$
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(e)}$