I am not sure if I understand proving results regarding unindexed families of sets and would appreciate some help.

(i) Suppose that $A$ is a set and $F$ is a family of sets.Prove that $A$\ $\bigcup$$F=\bigcap${$A$\ $B:B \in F $}.

My attempt:

$x \in A$\ $\bigcup F$

  • iff $x \in A$ and $x \notin \bigcup F$
  • iff $x \in A$ and $x \notin B $ for every $B \in F$.
  • iff $x \in A$\ $B$ for every $B \in F$
  • iff $x \in \bigcap A$\ $B$
  • iff $x \in$ $\bigcap${$A$\ $B:B \in F $}

Therefore : $A$\ $\bigcup$$F=\bigcap${$A$\ $B:B \in F $}.

(ii) Let $F$ and $G$ be two families of sets.Prove that $\bigcup(F\cup G)=(\bigcup F)\cup(\bigcup G)$

My attempt:

$x \in(F\cup G)$

  • iff $x\in (F \cup G)$ for some $F \in A$ and $F \in B$
  • iff $x \in F$ or $x \in G$ for some $F \in A$ and $F \in B$
  • iff $x \in \bigcup F$ or $x \in \bigcup G$
  • iff $x \in (\bigcup F) \cup (\bigcup G)$

Therefore: $\bigcup(F\cup G)=(\bigcup F)\cup(\bigcup G)$ with $A$ and $B $ being some families?? Does seperating $F \in A$ and $F \in B$ make sense?..originally i thought i should have done :$(F \cup G) \in A$..would that have been wrong?

Thank you for your time.


In your answer on (i) the fourth bullet is wrong and should be left out or interchanged with:

  • iff $x\in c$ for every $c\in\{A\setminus B\mid B\in F\}$

By answering (ii) two families/sets $A$ and $B$ "fall from the sky".

They are not mentioned in what you are asked to prove.

Equivalent are:

  • $x\in\bigcup(F\cup G)$
  • $x\in a$ for some $a\in F\cup G$
  • $x\in a$ for some $a$ that satisfies $a\in F$ or satisfies $a\in G$
  • $x\in a$ for some $a\in F$ or $x\in a$ for some $a\in G$
  • $x\in\bigcup F$ or $x\in\bigcup G$
  • $x\in\left(\bigcup F\right)\cup\left(\bigcup G\right)$

In this answer I stay in line with the notation that you practicize, but that is not how I would do it.

For me personally if $A$ is a set then $\cup A$ is again a set and this with: $$x\in\cup A\iff x\in a\text{ for some }a\in A$$ using the small cup. In that sense $\cup$ is an operator on sets.

Further $F\cup G$ is then an abbreviation of $\cup\{F,G\}$ and $\bigcup_{\lambda\in\Lambda}A_{\lambda}$ (using bigcup) is an abbreviation of $\cup\{A_{\lambda}\mid\lambda\in\Lambda\}$.

Similar story for cap and bigcap

  • $\begingroup$ ok, thank you ..i think i understand my mistake $\endgroup$ – HalfAFoot Mar 15 at 10:24
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Mar 15 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.