Union of subsets is a subset of the union

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Given two arbitrary families of sets, [ilmath]\{A_\alpha\}_{\alpha\in I} [/ilmath] and [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] such that [ilmath]\forall\alpha\in I[B_\alpha\subseteq A_\alpha][/ilmath] we have[1]:
  • [math]\bigcup_{\alpha\in I}B_\alpha\subseteq\bigcup_{\alpha\in I}A_\alpha[/math]

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


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

Let {{M|x\in\bigcup_{\alpha\in I}B_\alpha} this means
  • [ilmath]\exists\beta\in I[x\in B_\beta][/ilmath], let [ilmath]\beta[/ilmath] be defined as such.
By hypothesis, [ilmath]\forall\alpha\in I[B_\alpha\subseteq A_\alpha][/ilmath], again using the implies-subset relation we see:
  • [ilmath]x\in B_\beta\implies x\in A_\beta[/ilmath]
So we have [ilmath]x\in A_\beta[/ilmath]
Recall that [ilmath]x\in\bigcup_{\alpha\in I}A_\alpha\iff\exists\beta\in I[x\in A_\beta][/ilmath]
  • We have exactly the right side of this, so we also have
[ilmath]x\in\bigcup_{\alpha\in I}A_\alpha[/ilmath]
We have shown [ilmath]x\in\bigcup_{\alpha\in I}B_\alpha\implies x\in\bigcup_{\alpha\in I}A_\alpha[/ilmath], which (again, by the implies-subset relation) is exactly:
  • [ilmath]\bigcup_{\alpha\in I}B_\alpha\subseteq\bigcup_{\alpha\in I}A_\alpha[/ilmath] - As required.

This completes the proof.

See also


  1. Alec's (my) own work