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Show that if $A$ and $B$ are sets, then $(A\cap B) \cup (A\cap \overline{B})=A$.

So I have to show that $(A\cap B) \cup (A\cap \overline{B})\subseteq A$ and that $A \subseteq(A\cap B) \cup (A\cap \overline{B})$.

Lets begin with the first one:

If $x \in (A \cap B)$ it means $x \in A \wedge x \in B$.

If $x \in (A \cap \overline{B})$ it means $x \in A \wedge x \in \overline{B}$.

And the second one:

If $x \in A$ it means $x \in (A \cap B)$.

But here after I am confused.

Daniel W. Farlow
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Alim Teacher
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  • I would start with the definition of subset and show that both the left is a subset of the right and the right is a subset of the left, hence they are equal. Notice on the left either $x\in A$ or $x\in A$. – Todd Wilcox Apr 09 '15 at 20:39
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    Note that an element-chasing proof is really unnecessary here (hence my one-liner answer). Chasing elements and showing both directions seems like a lot of bang for almost no buck. Is this a homework problem and you're required to do an element-chasing proof? If not, I would definitely just use algebra of sets and use my answer. Otherwise, follow Ross's outline and complete your proof. – Daniel W. Farlow Apr 09 '15 at 20:45

3 Answers3

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Hint: $(A\cap B)\cup(A\cap\overline{B})=A\cap(B\cup\overline{B})=A$.

Daniel W. Farlow
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  • This is basically a full proof. – A. Thomas Yerger Apr 09 '15 at 20:46
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    @AlfredYerger I'd argue that it is a full proof. :p My point in marking it "hint" was ultimately to leave out the facts that I'm using distributivity and that $A$ is contained in $B\cup\overline{B}$...but now maybe OP will see this comment and the game will be up and no longer a hint in any meaning of the word. :/ – Daniel W. Farlow Apr 09 '15 at 20:48
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Following your approach of chasing elements, you can say If $x \in (A \cap \overline{B})$ it means $x \in A \wedge x \in \overline{B}$, so $x \in A \wedge x \not \in {B}$. Therefore $x \in ((A\cap B) \cup (A\cap \overline{B}))$ means $(x \in A \wedge x \in {B})\vee (x \in A \wedge x \not \in {B})$ and use the distributive principle

Ross Millikan
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Let $x\in (A \cap B) \cup (A \cap B^c)$. Then either $x\in (A \cap B) $ or $x\in (A \cap B^c) $. In the first case, $x\in A$ and $x\in B$, so $x\in A$. In the second case, $x\in A$ and $x\in B^c$, so $x\in A$. Thus $ (A \cap B) \cup (A \cap B^c)\subseteq A$.

Conversely, let $x\in A$. Then either $x\in B$ or $x\not\in B$. If $x\in B$, then $x\in (A \cap B)$, so $x\in (A \cap B) \cup (A \cap B^c)$. If $x\not\in B$, then $x\in B^c$. Thus $x\in (A \cap B^c)$ and $x\in (A \cap B) \cup (A \cap B^c)$. Thus $A \subseteq (A \cap B) \cup (A \cap B^c)$.

Putting these together gives $ (A \cap B) \cup (A \cap B^c)= A$.

  • Not sure if you know this which is why I'm commenting (if you do know, let me know and I'll delete)--you clearly see $\overline{B}$ and $B^c$ indicate that same thing. OP may not have seen this notation and/or maybe you didn't know \overline{B} produced $\overline{B}$ perhaps? Nothing important. :) – Daniel W. Farlow Apr 09 '15 at 20:52
  • \overline{B} is so much more to type than B^c :) –  Apr 09 '15 at 20:54
  • So which one is more correct Ross's outline or this? – Alim Teacher Apr 09 '15 at 21:06
  • @AlimTeacher Do you have to use an element-chasing proof? If so, then read avid69's and Ross' answers more closely. – Daniel W. Farlow Apr 09 '15 at 21:38