Union of subsets is a subset of the union

Statement

Given two arbitrary families of sets, $\{A_\alpha\}_{\alpha\in I}$ and $\{B_\alpha\}_{\alpha\in I}$ such that $\forall\alpha\in I[B_\alpha\subseteq A_\alpha]$ we have:

• $$\bigcup_{\alpha\in I}B_\alpha\subseteq\bigcup_{\alpha\in I}A_\alpha$$

This seems quite trivial (and it is) but it is a very useful result

Proof

We will show that $\bigcup_{\alpha\in I}B_\alpha\subseteq\bigcup_{\alpha\in I}A_\alpha$ by using the implies-subset relation, and showing $x\in\bigcup_{\alpha\in I}B_\alpha\implies x\in\bigcup_{\alpha\in I}A_\alpha$

Let {{M|x\in\bigcup_{\alpha\in I}B_\alpha} this means

• $\exists\beta\in I[x\in B_\beta]$, let $\beta$ be defined as such.

By hypothesis, $\forall\alpha\in I[B_\alpha\subseteq A_\alpha]$, again using the implies-subset relation we see:

• $x\in B_\beta\implies x\in A_\beta$

So we have $x\in A_\beta$

Recall that $x\in\bigcup_{\alpha\in I}A_\alpha\iff\exists\beta\in I[x\in A_\beta]$

• We have exactly the right side of this, so we also have

$x\in\bigcup_{\alpha\in I}A_\alpha$

We have shown $x\in\bigcup_{\alpha\in I}B_\alpha\implies x\in\bigcup_{\alpha\in I}A_\alpha$, which (again, by the implies-subset relation) is exactly:

• $\bigcup_{\alpha\in I}B_\alpha\subseteq\bigcup_{\alpha\in I}A_\alpha$ - As required.

This completes the proof.

See also

• The intersection of sets is a subset of each set