Set Theory question show that

Question

Would this be correct

Let $A$ and $X$ be subsets of a universe $U$. Show that if

$A \cup X =$ Universe
$A \cap X = \emptyset$ then $X=A^c$

$A \cup X =$ Universe (given)
$A \cup A^c =$ Universe (complement law)
$A \cap X = \emptyset$ (given)
$A \cap A^c = \emptyset$ (complement law)

therefore:

$X = A^c$

No, you need to add more details even though they seem obvious to you. Prove that $X \subset A^c$ and $A^c \subset X$ — dani_s, Feb 25 '14 at 22:12
First, welcome to the stack exchange! I am guessing this is an intro to proofs class, and for that reason, I am guessing they want you to do a double subset inclusion proof as dani points out. — Ryan Sullivant, Feb 25 '14 at 22:12
By the way, it's "complement", and not "compliment". A complement completes something: $A\cup A^{c}$ is the universe. A compliment is when you say something nice. — Unwisdom, Feb 25 '14 at 22:23
@RyanSullivant Thanks man. This stuff is pretty difficult. Can you recommend any good resources? — Michael James, Feb 26 '14 at 09:02
@MichaelJames Sure, no worries. My advice is just stick with it and don't get down on yourself; eventually it will click. Mathematics is something that has to internalize in us, and often that process can take much longer than we would like. As for resources, I have How to Prove It by Velleman, (although I think I learned from lecture notes). Also, see this: http://math.stackexchange.com/questions/190981/how-to-be-good-at-proving/190987#190987 Asking questions is always good, so you've started down the right path! — Ryan Sullivant, Feb 26 '14 at 23:46
@MichaelJames It may also be worth it to take a look at this course: https://www.coursera.org/course/maththink — Ryan Sullivant, Feb 26 '14 at 23:47

score 1 · Accepted Answer · answered Feb 25 '14 at 22:16

There are lots of different inference schemes for this sort of thing, but I don't see exactly how your conclusion follows. There are, however, two general approaches. One is algebraic, treating the collection of subsets of some universe as a Boolean algebra, and using the Boolean algebra axioms to conduct inference. The other is usually a little easier for beginners, and is more element based, focusing on the "$\in$" relationship.

Under this second approach the easiest way to show equality of sets is to appeal to the Axiom of Extensionality: $$C=D \iff \big( x\in C \iff x\in D\big).$$ The simpler form is this: $$C=D \iff \big( (C\subseteq D) \,\&\, (D\subseteq C)\big).$$

In our case, we want to show that $X=A^{c}$. I'll use extensionality, since it's faster and I have to rush to get my bus.

Suppose that $x\in X$. Then since $X\cap A=\emptyset$, it must be the case that $x\notin A$. Thus $X\subseteq A^{c}$. Now take $x\in A^{c}$. Since $x\in X\cup A$, we must have $x\in X$. Thus $A^{c}\subseteq X$, as required.

score 0 · Answer 2 · answered Feb 26 '14 at 07:15

Letting $\;x\;$ range over $\;U\;$, we can just expand the definitions, and calculate: \begin{align} & A \cup X = U \;\land\; A \cap X = \emptyset \\ \equiv & \qquad \text{"set extensionality, twice"} \\ & \langle \forall x :: x \in A \cup X \;\equiv\; x \in U \rangle \;\land\; \langle \forall x :: x \in A \cap X \;\equiv\; x \in \emptyset \rangle \\ \equiv & \qquad \text{"definitions of $\;\cup, U, \cap, \emptyset\;$"} \\ & \langle \forall x :: x \in A \lor x \in X \;\equiv\; \text{true} \rangle \;\land\; \langle \forall x :: x \in A \land x \in X \;\equiv\; \text{false} \rangle \\ \equiv & \qquad \text{"logic: simplify, DeMorgan, merge $\;\forall\;$-quantifications"} \\ & \langle \forall x :: (x \in A \lor x \in X) \;\land\; (x \not\in A \lor x \not\in X) \rangle \\ \equiv & \qquad \text{"logic: rewrite as $\lnot p \lor q\;$ as $\;p \Rightarrow q\;$, twice"} \\ & \langle \forall x :: (x \not\in A \Rightarrow x \in X) \;\land\; (x \in X \Rightarrow x \not\in A) \rangle \\ \equiv & \qquad \text{"logic: bidirectional"} \\ & \langle \forall x :: x \in X \;\equiv\; x \not\in A \rangle \\ \equiv & \qquad \text{"definition of $\;^x\;$; set extensionality"} \\ & X = A^c \\ \end{align}

Incidentally, this also proves that (1) and (2) follow from $\;X = A^c\;$.

Set Theory question show that

2 Answers2