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For any subsets of $\mathbb{Z}$ named $X,Y$, we define $+$ operation as below: $$X+Y = \{z \in \mathbb{Z}\mid z=x+y \:\text{ for some } x \in X, y \in Y\}.$$ Suppose that $\mathbb{Z}= A \cup B$ where $A$ and $B$ are disjoint, $0 \in A$, and $B$ is a nonempty set. Also suppose that $A+A$, $A+B$ and $B+B$ are each equal to either $A$ or $B$. Can you get the conclusion that $A$ must be set of all even integers and $B$ must be the set of all odd integers?

Dylan Moreland
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Shannon
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  • When you define "+", did you mean to define it in the usual way (i.e. {z|z=x+y for x in X, y in Y}) or was that not a typo? – Lopsy Feb 03 '12 at 02:13
  • That "+" is the operation between two sets X + Y. I may use another simple to distinguish it from the usual one. Sorry. – Shannon Feb 03 '12 at 02:16
  • @Lopsy, yayaya, you are right. Its z=x+y – Shannon Feb 03 '12 at 02:19

1 Answers1

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Yes. Since $0\in A$, we have $A\subseteq A+A$ and $B\subseteq A+B$, so it follows that $A+A=A$ and $A+B=B$. Let $x$ be an element of $A$. If $-x$ were in $B$, then we would have $0=x+(-x)\in x+B\subseteq A+ B=B$ which contradicts $0\in A$ and $A\cap B=\emptyset$. Thus $-x$ must be in $A$.

Since $A$ is a nonempty subset of $\mathbb Z$ that is closed under addition and additive inverses, $A$ is a subgroup of $\mathbb Z$. Therefore $A=k\mathbb Z$ for some $k\geq0$. We can rule out $k=0$, because if $B=\mathbb Z\setminus\{0\}$, then $B+B=\mathbb Z$. We can rule out $k=1$, because $B$ is nonempty. Finally, we can rule out $k>2$, because if that were the case, $1+1$ would be in $B+B$ and in $B$, while $1-1$ would be in $B+B$ and in $A$, contradicting the hypothesis that $B+B$ is one of $A$ or $B$.

Therefore $A=2\mathbb Z$ is the set of even numbers, and $B=\mathbb Z\setminus 2\mathbb Z=1+(2\mathbb Z)$ is the set of odd numbers.

Here is an alternative to the approach in the second paragraph, not referring to groups. Picking up from the first paragraph, so far we know that $A+A=A$, $A+B=B$, $-A=A$. We can see similarly that $-B=B$ (if $x$ were in $B$ with $-x\in A$, we would have $0=x-x\in A+B=B$ again). Since $B$ is nonempty, there exists $x\in B$, and therefore $0=x-x\in B+B$. This implies that $B+B=A$.

If $1$ were in $A$, then using $A+A=A$ and induction we would have all positive integers in $A$, and using $-A=A$ we would have all the rest. Since $B$ is nonempty, this is impossible, so $1$ is in $B$. Thus $2=1+1\in B+B=A$. Using $A+A=A$ and induction, all positive even integers are in $A$, and using $-A=A$ we get all the rest. So $2\mathbb Z\subseteq A$. Therefore $B=A+B\supseteq (2\mathbb Z)+1$, which is the set of odd numbers. Since $A$ contains all even numbers and $B$ contains all odd numbers, we in fact have $A=2\mathbb Z$ and $B=1+(2\mathbb Z)$.

(Actually, this second approach seems somewhat similar to the approach in a now deleted answer.)

Jonas Meyer
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  • Thanks for your help! I have two questions for you. First, I don't get why -x must be in A. Second, I dont know the notation of a set \ a set. – Shannon Feb 03 '12 at 03:03
  • For your first question, we know that if $-x\in B$, then $0=x+(-x)\in x+ B\subseteq A+B= B$, but this contradicts $0\in A$ and $A\cap B=\emptyset$. For your second question, $X\setminus Y$ is notation meaning ${x\in X: x\not\in Y}$, sometimes called the set-theoretic difference or relative complement of $Y$ in $X$. E.g., $\mathbb Z\setminus {0}$ is the set of nonzero integers. In this problem we know $A=\mathbb Z\setminus B$ and $B=\mathbb Z\setminus A$. A reference is the following article: http://en.wikipedia.org/wiki/Complement_%28set_theory%29#Relative_complement – Jonas Meyer Feb 03 '12 at 03:07
  • @Shannon: I edited the first paragraph to clarify. – Jonas Meyer Feb 03 '12 at 03:22
  • I am still confused. If B=Z{0}, then B+B is a set without 0 in it. So it doesnt equal Z. – Shannon Feb 03 '12 at 03:34
  • @Shannon: $1$ and $-1$ are both in $\mathbb Z\setminus {0}$, so $0=1+(-1)$ is in $(\mathbb Z\setminus{0})+(\mathbb Z\setminus{0})$. For all $k\in\mathbb Z$, you can always find two nonzero integers whose sum is $k$, and therefore $(\mathbb Z\setminus{0})+(\mathbb Z\setminus{0})=\mathbb Z$. – Jonas Meyer Feb 03 '12 at 03:40
  • Sorry, I have another question. How do you know A=kZ? A could be just a random subset of Z. – Shannon Feb 03 '12 at 04:17
  • @Shannon: There I used the fact that the only subgroups of the integers with addition are the cyclic subgroups. I did not know whether this would be familiar to you, so I'm glad you asked. It is a standard result usually covered in introductory courses in group theory that every subgroup of a cyclic group is cyclic. There was a question about proving it here: http://math.stackexchange.com/questions/6998. It is possible to show without using the group theoretic language, as it really only relies on division with remainder for integers. ($\mathbb Z$ is generated as a group by $1$.) – Jonas Meyer Feb 03 '12 at 04:27
  • Perfect. Group theory, that's what I am working on it. Appreciate for your help sooooo much. – Shannon Feb 03 '12 at 04:37
  • @Shannon: You're welcome. I added another approach that doesn't use that. – Jonas Meyer Feb 03 '12 at 04:49