1
$\begingroup$

Let U be the universe and let A be a subset of U. Then prove:

a. $A$ $\cup$ $A^c$= $U$

b. $A$ $\cap$ $A^c$=$\emptyset$

Proof:

a. Let $x$ $\in$ $A$ $\cup$ $A^c$ $\Rightarrow$ $x$ $\in$ $U$

Then $x$ $\in$ $A$ or $x$ $\in$ $A^c$ $\Rightarrow$ $x$ $\in$ $U$

By definition, the complement of a set $A$ is $A^c$ = $U$ $-$ $A$ where $x$ $\in$ U and $x$ $\notin$ $A$.

It follows that $x$ $\in$ $A$ or $x$ $\in$ U and $x$ $\notin$ $A$ $\Rightarrow$ $x$ $\in$ $U$

Thus it is proven that $x$ $\in$ $U$ and $A$ $\cup$ $A^c$= $U$.

$\blacksquare$

b. Let $x$ $\in$ $A$ $\cap$ $A^c$ $\Rightarrow$ $x$ $\in$ $\emptyset$

Then $x$ $\in$ $A$ and $x$ $\in$ $A^c$ $\Rightarrow$ $x$ $\in$ $\emptyset$

By definition of the complement of A, it follows that $x$ $\notin$ $A$.

The intersection is empty and therefore $x$ $\in$ $\emptyset$.

$\blacksquare$

These are my proofs. If someone could looks over them to see if I missed anything. That'd be greatly appreciated.

$\endgroup$
  • $\begingroup$ It looks quite messy to me. You seem to have the correct ideas, but your formatting is strange. You have "$\implies x\in U$" appearing at the end of almost every line for your proof for $a$, and it seems as though this is just a constant reminder to yourself that you are wishing to reach the end goal of $x\in U$ rather than an actual step in the proof. The same awkwardness occurs in your part (b). $\endgroup$ – JMoravitz Sep 21 '17 at 21:08
  • $\begingroup$ Perhaps, @JMoravitz, but it certainly shows involvement in the question, effort, thought, etc. So, since I'm in a happy mode right now, kudos to the asker, ErinA! $\endgroup$ – amWhy Sep 21 '17 at 22:20
2
$\begingroup$

It seems like you're on the right way, but I think you're confusing different notations or concepts. At the end of (b), you claim $x \in \emptyset$, but $\emptyset$ does not contain any elements.

The definitions of complements, unions ($\cup$) and intersection ($\cap$) give us:

$A^c = \{x: x\notin A\} $

$A \cup B = \{x:x \in A \; or \; x \in B\}$ (both may hold)

$A \cap B = \{x:x \in A \; and \; x \in B \}$.

In (a) you have $A \cup A^c = \{x:x \in A \;\; or \;\; x \in A^c \}$. If we take any $x \in U$, one of these must hold (by the law of excluded middle). Therefore, $A \cup A^c = U$.

In (b), what you want to do is prove that there are no elements in $A \cap A^C$. Using the definition, you have $A \cap A^c = \{x:x \in A \;\; and \;\; x \in A^c \}$, which does not hold for any $x$ (by the law of non-contradiction). Hence, $A \cap A^c = \emptyset$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.