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

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

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


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


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 $x$ $\in$ $\emptyset$.


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

  • $\begingroup$ 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

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


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