Sets and Venn diagrams

• February 18th 2014, 01:46 AM
vanilla5085
Sets and Venn diagrams
Hi,I was just wondering if anyone could help me with this question?Simplify:A∪A′ for any set A∈U
• February 18th 2014, 02:20 AM
Plato
Re: Sets and Venn diagrams
Quote:

Originally Posted by vanilla5085
Hi,I was just wondering if anyone could help me with this question?Simplify:A∪A′ for any set A∈U

For any $x\in\mathcal{U}$ it is true that $x\in A\text{ or }x\notin A$ so $A\cup A'=~?$
• February 18th 2014, 03:59 AM
Hartlw
Re: Sets and Venn diagrams
U, by definition of A'.
• February 18th 2014, 09:37 PM
vanilla5085
Re: Sets and Venn diagrams
Thanks Plato and Hartlw,
the answer is U but I don't understand why.
• February 18th 2014, 09:53 PM
Prove It
Re: Sets and Venn diagrams
A U A' means "the element chosen could come from either set 'A' or set 'not A' or both". Since it's impossible to be from both, that means it has to either come from A or not come from it. Thus it could be anything in the universal set.
• February 19th 2014, 05:12 AM
Hartlw
Re: Sets and Venn diagrams
A' is all elements in U not in A. If you add A you get U.
• February 19th 2014, 05:29 AM
SlipEternal
Re: Sets and Venn diagrams
First show that $A\cup A' \subseteq U$. Then show $U \subseteq A \cup A'$.

Claim: $A \cup A' \subseteq U$

Proof:
Since $A \subseteq U$ so $A' \subseteq U$. Hence, given any $x \in A \cup A'$, at least one of the following is true: $x \in A$ or $x \in A'$. In either case, by the definition of subset, $x \in U$, so $A \cup A' \subseteq U$ as claimed.

Claim: $U \subseteq A \cup A'$

Proof: See post #2 (from Plato) or #5 (from Prove It)
• February 19th 2014, 05:44 AM
Hartlw
Re: Sets and Venn diagrams
Quote:

Originally Posted by Hartlw
A' is all elements in U not in A. If you add A you get U.

That is perfectly clear and correct, based on definition of A'. The abstractions are redundant.
• February 21st 2014, 01:41 AM
vanilla5085
Re: Sets and Venn diagrams
Thanks everyone!