Prove union and intersection of a set with itself equals the set

by RoRi

June 20, 2015

Prove that $A \cup A = A$ and $A \cap A = A$ .

Proof. Let $x$ be an arbitrary element of $A \cup A$ . Then $x \in A$ or $x \in A$ ; hence, $x \in A$ . Thus, $A \cup A \subseteq A$ .
Conversely, if $x$ is an arbitrary element of $A$ then $x \in A \cup A$ since it is in $A$ . Thus, $A \subseteq A \cup A$ .
Therefore, $A \cup A = A. \qquad \blacksquare$

Proof. Let $x$ be an arbitrary element of $A \cap A$ . Then $x \in A$ and $x \in A$ ; hence, $x \in A$ . So, $A \cap A \subseteq A$ .
Conversely, if $x \in A$ is arbitrary, then $x \in A$ and $x \in A$ ; hence, $x \in A \cap A$ . Thus, $A \cap A \subseteq A$ .
Therefore, $A \cap A = A. \qquad \blacksquare$

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3 comments

May 18, 2019 at 10:11 am

Manik Sharma says:

On the top write in the last set itself

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September 19, 2017 at 12:16 am

Jolene Hunt says:

Hi there,

You listed Lara Alcock’s book, but misspelled her name as Laura in the link. Thanks for the recommendation though :)

Reply
July 20, 2016 at 2:44 am

David He says:

Hey Rori,

Should “A \cap A \subseteq A” on the second proof be reversed?

Best,

David

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