Cardinality of Subset Relation on Power Set of Finite Set

Theorem

Let $S$ be a set such that:

$\left|{S}\right| = n$

where $\left|{S}\right|$ denotes the cardinality of $S$.

From Subset Relation on Power Set is Partial Ordering we have that $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is an ordered set.

The cardinality of $\subseteq$ as a relation is $3^n$.

Proof

Let $X \in \mathcal P \left({S}\right)$.

Since $X \subseteq S$, it follows that:

$X' \subseteq X \implies X' \in \mathcal P \left({S}\right)$

because the Subset Relation is Transitive.

From Cardinality of Power Set of Finite Set, it follows that for any $X \in \mathcal P \left({S}\right)$:

$\left\{{X' \in \mathcal P \left({S}\right): X' \subseteq X}\right\}$

has $2^{\left|{X}\right|}$ elements.

Therefore, the cardinality of $\subseteq$ is given by:

$\displaystyle \sum_{X \mathop\subseteq S} 2^{\left|{X}\right|}$

Let us split the sum over $\left|{X}\right|$:

$\displaystyle \sum_{X \mathop\subseteq S} 2^{\left|{X}\right|} = \sum_{k \mathop = 0}^n \sum_{\substack{X \mathop\subseteq S \\ \left|{X}\right| = n}} 2^{\left|{X}\right|}$

It now follows from Cardinality of Set of Subsets that:

$\displaystyle \left|{\subseteq}\right| = \sum_{k \mathop = 0}^n \binom n k 2^k$

From the Binomial Theorem:

$\displaystyle \sum_{k \mathop = 0}^n \binom n k 2^k = \left({1 + 2}\right)^n$

Hence:

$\displaystyle \left|{\subseteq}\right| = 3^n$

$\blacksquare$

Sources

• 1983: George F. Simmons: Introduction to Topology and Modern Analysis ... (previous): $\S 1$: Sets and Set Inclusion