Union of Blocks is Set of Points

Theorem
Let $(X, \mathcal B)$ be a pairwise balanced design.

That is, let $(X,\mathcal B)$ be a design, with $|X| \ge 2$, and the number of occurrences of each pair of distinct points in $\mathcal B$ be $\lambda$ for some $\lambda > 0$ constant.

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

Lemma
Let $(X, \mathcal B)$ be a balanced incomplete block design.

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

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

Let $\mathcal B = \{\!\{y_1,y_2,\ldots, y_b\}\!\}$.

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

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

$Y \subseteq X$:
By definition, $\mathcal B$ is a multiset of subsets of $X$.

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

By Union of Subsets is Subset, $Y$ itself is a subset of $\mathcal B$

$X \subseteq Y$:
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 $|X|\ge 2$.

Then $\{x_i,x_j\} \subseteq Y$ 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.

Proof of Lemma
By the definition of balanced incomplete block design, $(X,\mathcal B)$ is itself a type of pairwise balanced design.

Thus the lemma follows from the main result.