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Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition.

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It's actually a practice test. My teacher does not give us homework or many examples but mostly lectures and I am having a very hard time figuring out how to solve problems in set theory. I am getting used to reading and writing it, but actually using it is proving to be a greater challenge than I expected it to be. – Omar Nov 25 '12 at 19:18
up vote 5 down vote accepted

First let us recall Zorn's lemma.

Zorn's lemma. Suppose that $(P,\leq)$ is a non-empty partial ordered set such that whenever $C\subseteq P$ is a chain, then there is $p\in P$ that for every $c\in C$, $c\leq p$. Then $(P,\leq)$ has a maximal element.

To use Zorn's lemma, if so, one has to find a partial order with the above property (every chain has an upper bound) an utilize the maximality to prove what is needed.

We shall use the partial order whose members are $(A,B)$ where $A,B$ are disjoint subsets of $\mathbb R^+$ each is closed under addition. We will say that $(A,B)\leq (A',B')$ if $A\subseteq A'$ and $B\subseteq B'$.

This is obviously a partial order. It is non-empty because we can take $A=\mathbb N\setminus\{0\}$ and $B=\{n\cdot\pi\mid n\in\mathbb N\setminus\{0\}\}$, both are clearly closed under addition and disjoint.

Suppose that $C=\{(A_i,B_i)\mid i\in I\}$ is a chain, let $A=\bigcup_{i\in I}A_i$ and $B=\bigcup_{i\in I} B_i$. To see that these sets are disjoint suppose $x\in A\cap B$ then for some $A_i$ and $B_j$ we have $x\in A_i\cap B_j$. Without loss of generality $i<j$ then $x\in A_j\cap B_j$ contradiction the assumption that $(A_j,B_j)\in P$ and therefore these are disjoint sets. The proof that $A$ and $B$ are closed under unions is similar.

Then $(A,B)\in P$ and therefore is an upper bound of $C$. So every chain has an upper bound and Zorn's lemma says that there is some $(X,Y)$ which is a maximal element.

Now all that is left is to show that $X\cup Y=\mathbb R^+$. Suppose that it wasn't then there was some $r\in\mathbb R^+$ which was neither in $X$ nor in $Y$, then we can take $X'$ to be the closure of $X\cup\{r\}$ under addition. If $X'\cap Y=\varnothing$ then $(X',Y)\in P$ and it is strictly above $(X,Y)$ which is a contradiction to the maximality. Therefore $X'\cap Y$ is non-empty, but then taking $Y'$ to be the closure of $Y\cup\{r\}$ under addition has to be disjoint from $X$, and the maximality argument holds again.

In either case we have that $X\cup Y=\mathbb R^+$.

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The part showing that $Y'\cap X=\varnothing$ is a bit tricky, but if you think about it you can make it on your own (hint, argue by contradiction that otherwise you get that for some $n>1$, $nr+x\in Y$ and $nr+y\in X$. Deduce that $x+y\in X\cap Y$). – Asaf Karagila Nov 25 '12 at 19:38

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