# SET Operations

• Apr 17th 2010, 04:33 AM
npm1
SET Operations
hi all,,,
i need a hand on sumin.....
Use the laws of algebra of sets to show that (X union Y) intersection X = X
please specify which rules you use in the derivation
(Hint: you may use the fact that (X union Y) intersection X = (X union Y) intersection (X union empty Set) [rule 5a] a sequence of identities).
• Apr 17th 2010, 07:17 AM
Failure
Quote:

Originally Posted by npm1
hi all,,,
i need a hand on sumin.....
Use the laws of algebra of sets to show that (X union Y) intersection X = X
please specify which rules you use in the derivation
(Hint: you may use the fact that (X union Y) intersection X = (X union Y) intersection (X union empty Set) [rule 5a] a sequence of identities).

$(X\cup Y)\cap X = (X\cup Y)\cap (X\cup \emptyset) = X\cup (Y\cap \emptyset) = X\cup \emptyset = X$
• Apr 17th 2010, 07:37 AM
npm1
can u explain to me please
can u explain your awnser to me..... i don't under stand
• Apr 17th 2010, 08:49 AM
Failure
Quote:

Originally Posted by npm1
can u explain your awnser to me..... i don't under stand

An explanation always has to refer to what the person, who is to receive the explanation, already knows. Since I don't know exactly what you already know, it's going to be guesswork on my part, but I'll try to improve somewhat on my first attempt in this direction

Let me annotate the various equality signs with numbers, so that I can refer to them with precision:
$(X\cup Y)\cap X \overset{\text{(1)}}{=} (X\cup Y)\cap (X\cup \emptyset) \overset{\text{(2)}}{=} X\cup (Y\cap \emptyset) \overset{\text{(3)}}{=} X\cup \emptyset \overset{\text{(4)}}{=} X$

Now the explanation for why the corresponding equality sign holds (if the above sequence of transformations is read from left to right):

(1) is an application of the hint that you were given in the problem statement. This is one of the two "absorption laws, namely that $X\cup \emptyset=X$ is true for any set $X$. Thus we are allowed to replace any occurrence of $X$ with $X\cup \emptyset$, if we want to do that.

(2) is an application of the distributive law for $\cap$ and $\cup$. In fact, there are two distributive laws, one of which is $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$ which was applied here (in the right-to-left direction, by setting $A := X, B := Y, C :=\emptyset$).
Btw: The other distributive law is $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$, but we didn't need it here.

(3) is is an application of the other "absorption law", which says that $Y\cap \emptyset = \emptyset$, for any set $Y$ is true. Thus we can replace any occurrence of $Y\cap \emptyset$ by $\emptyset$, if we want to do that.

(4) This is yet another application of the first absorption law, already used in (1), namely that $X\cup\emptyset=X$, for any set $X$ is true.