# Thread: Set theory "basic" identities question

1. ## Set theory "basic" identities question

these things are confusing, please help me fill out the blanks and correct me :
(u - is the universal set)

1. $\displaystyle \emptyset \cup \emptyset = \emptyset$ ?

2. $\displaystyle \emptyset \cap \emptyset = u$ ?

3. $\displaystyle u \cup u =$ ?

4. $\displaystyle u \cap u =$ ?

5. $\displaystyle u \cup \emptyset = u$ ?

6. $\displaystyle u \cap \emptyset = \emptyset$ ?

* - $\displaystyle if B\setminus A=A^C$ does it mean that B=u ?

thanks.

2. Originally Posted by Laban
(u - is the universal set)

2. $\displaystyle \emptyset \cap \emptyset = u$ ?

3. $\displaystyle u \cup u =$ ?

4. $\displaystyle u \cap u =$ ?

* - $\displaystyle if B\setminus A=A^C$ does it mean that B=u ?
2. $\displaystyle \emptyset \cap \emptyset = \emptyset$.

3. $\displaystyle u \cup u = U$ .

4. $\displaystyle u \cap u = U$ .

* - $\displaystyle if B\setminus A=A^c$ it means that $\displaystyle A^c\subseteq B.$

3. thanks!

regarding the 2nd question I found this : empty sets union & intersection
is it not correct? or maybe I've misinterpreted something?

and if $\displaystyle u\setminus A=A^C$ and you get this situation $\displaystyle B\setminus A=A^C$ how come b≠u?

4. First, lets be absolutely clear that you said absolutely nothing about dealing with families of sets.
The link you provide is discussing union and intersections of families of sets.
It is easy to get different mathematicians into an argument over this question.
Herbert Enderton remarks that trying to define $\displaystyle \bigcap \emptyset$ is like division by zero it is best to say it is undefined. But he adds that logicians particularly do not like that. That is what you see at ask DrMath. In his text, Stoll restricts intersections of collections to non-empty collections.

In so called naïve set theory, we have this rule: $\displaystyle A\cap\emptyset=\emptyset$ for all $\displaystyle A$.
In that case, it follows that $\displaystyle \emptyset\cap\emptyset=\emptyset$.

As to your second question: $\displaystyle P \cap Q = Q\; \Rightarrow \;Q \subseteq P$.