# DeMorgan's Law regarding unindexed families of sets

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

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

(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.

## 1 Answer

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

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