Union of Blocks is Set of Points

Theorem

Let $\struct {X, \BB}$ be a pairwise balanced design.

That is, let $\struct {X, \BB}$ be a design, with $\size X \ge 2$, and the number of occurrences of each pair of distinct points in $\BB$ be $\lambda$ for some $\lambda > 0$ constant.

Then the set union of all the subset elements in $\BB$ is precisely $X$.

Corollary

Let $\struct {X, \BB}$ be a balanced incomplete block design.

Then the set union of all the subset elements in $\BB$ is precisely $X$.

Proof

Let $X = \set {x_1, x_2, \ldots, x_v}$.

Let $\BB = \multiset {y_1, y_2,\ldots, y_b}$, where the notation denotes a multiset.

Let $Y = \ds \bigcup_{i \mathop = 1}^b y_i$.

We shall show that $Y \subseteq X$ and $X \subseteq Y$.

$Y \subseteq X$:

By definition, $\BB$ is a multiset of subsets of $X$.

This means that each $y_i \in Y$ is a subset of $X$ for $i \in \closedint 1 b$.

By Union of Family of Subsets is Subset, $Y$ itself is a subset of $\BB$.

$\Box$

$X \subseteq Y$:

Let $x_i \in X$ be arbitrary.

Choose any $x_j \in X$ such that $i \ne j$.

Such an $x_j$ necessarily exists because by hypothesis $\card X \ge 2$.

Then $\set {x_i, x_j} \subseteq Y$ by the definition of blocks, as $\lambda > 0$ by hypothesis.

But by the definition of subset, this implies that $x_i$ is an element in $Y$.

Hence the result, as $x_i$ was arbitrary.

$\blacksquare$