# Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition.

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.

• 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, 2012 at 19:18

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) and 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 $$i then $$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 addition 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 $$X\cup Y=\mathbb R^+$$.

• 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 $x+y\in X\cap Y$). Nov 25, 2012 at 19:38
• you ought to be right as your answer was accepted, and I ought to figure the hint on my own, but: 1. I see $nr+x\in Y$ and $mr+y\in X$, don't see how we could assume $n=m$, and 2. say $nr+x=v\in Y$ and $nr+y=u\in X$, I get $x+y=u+v-2nr$, how does this help? Again, you ought to be right, I will try to figure it. Well, I made some progress, if $n=m$, i.e. if $nr+x=v\in Y$ and $nr+y=u\in X$ then $x+u=y+v\in X\cap Y$, a contradiction, now I need to figure why we could assume $n=m$. Did you really mean that $x+y\in X\cap Y$, and not $x+u$? May 17, 2017 at 23:57
• So here is how I fill in the missing part, and I wonder if I make it overly complicated. Assume that both $X'\cap Y\not=\emptyset$ and $X\cap Y'\not=\emptyset$. There is a smallest $n\ge1$ and some $x\in X,v\in Y$ with $nr+x=v$. There is a smallest $m\ge1$ and some $y\in Y,u\in X$ with $mr+y=u$. If $n=m$ done by my previous comment. Else wolog $n>m$, then $x+u+(n−m)r=y+v$ with $x+u\in X$, $y+v\in Y$, and $1\le n−m<n$ contradicting that $n$ was minimal. Am I missing something easier? Well, I am happy with this, doesn't look too long to me anymore, just writing the details. May 18, 2017 at 3:04
• @Mirko Another way to see this: if $x_0+nr=y_0$ and $y_1+mr=x_1$, then $mx_0+nx_1=my_0+ny_1$ is in $X\cap Y$. May 18, 2017 at 3:28

Pick a basis for $$\mathbb R$$ over $$\mathbb Q$$ and let $$v$$ be an element of the basis.

Each element in $$\mathbb R^+$$ can be expressed uniquely as a lineal combination of the basis (containing of course only a finite number of coefficients different from $$0$$). Let $$v$$ be an element of the basis.

Consider the set $$A$$ of elements of $$\mathbb R^+$$ in which the coefficient of $$v$$ is non-negative.

Consider the set $$B$$ of the elements of $$\mathbb R^+$$ in which the coefficient of $$v$$ is negative.

These sets do the trick.

• basis over what? May 17, 2017 at 23:22
• any proper subfield of $\mathbb R$ does the trick May 17, 2017 at 23:36

Here's a proof which uses Teichmüller–Tukey lemma:

If $\mathcal{F}\ne\emptyset$ is a family of finite character, i.e $X\in \mathcal{F} \iff$ every finite subset of $X$ is in $\mathcal{F}$, then $\mathcal{F}$ has a member which is maximal under inclusion.

This lemma is equivalent to Zorn's lemma. So not quite the exact requirement, but perhaps a simpler proof.

Now, let $\mathcal{F}$ be the collection of $A\subset \mathbb{R}^+$ such that if $x_1,...,x_n \in A$ (perhaps with repetitions) then $\sum_{i=1}^n x_i \notin \mathbb{N}$. Then $\mathcal{F}$ is of finite character, since the condition only requires finite subsets of each $A\in\mathcal{F}$. $\{\pi\}\in\mathcal{F}$ so it is not empty. Let $A$ be maximal in $\mathcal{F}$.

• $A$ is closed under addition since for $x,y\in A$, if $x+y\notin A$ by maximality it means that for some $x_1,...,x_n \in A$, $\sum_{i=1}^n x_i+(x+y) \in \mathbb{N}$ but this is still a finite sum of elements of $A$, contradiction.
• $\mathbb{R}^+\setminus A$ is closed under addition, since for $x,y\in \mathbb{R}^+\setminus A$, by maximality and closure under addition of $A$ there are $a,b\in A$ such that $a+x,b+y\in \mathbb{N}$, but then also $a+b+(x+y)\in\mathbb{N}$ so we can't have $x+y\in A$.

So, $(A,\mathbb{R}^+\setminus A)$ is the required partition.

• That doesn't answer the question of using Zorn's lemma specifically. Jul 10, 2018 at 10:33
• @HenningMakholm We'll yes, that's what I wrote. Since I saw an answer using the fact that every vector space has a basis, which is also a consequence Zorn's lemma, but not the lemma itself, I figured this would be ok as well. Jul 12, 2018 at 8:38

Let $$\mathcal{P}$$ the set of the disjoint pairs $$(A,B)$$, where $$A,B\subseteq\mathbb{R}^+$$ are not empty and each one is closed under addition and multiplication by a positive rational number. Note that $$\mathcal{P}\neq\emptyset$$ because if we consider $$X=\mathbb{Q}^+$$ and $$Y=\{n\sqrt{2}:n\in\mathbb{Q}^+\}$$, then $$(X,Y)\in\mathcal{P}$$. Define $$\leq$$ as follows:

$$(X_1,Y_1)\leq(X_2,Y_2)$$ if and only if $$X_1\subseteq X_2$$ and $$Y_1\subseteq Y_2$$, for all $$(X_1,Y_1),(X_2,Y_2)\in\mathcal{P}$$.

Clearly, $$(\mathcal{P},\leq)$$ is a partially ordered set. Furthermore, it is easy to see that every chain in $$(\mathcal{P},\leq)$$ has an upper bound. We can now apply Zorn's lemma, so $$(\mathcal{P},\leq)$$ has a maximal element. Let $$(A,B)\in\mathcal{P}$$ a maximal element of $$(\mathcal{P},\leq)$$. We only have to show that $$A\cup B=\mathbb{R}^+$$.

Suppose that $$\mathbb{R}^+\not\subseteq A\cup B$$. Therefore, there exists $$x\in\mathbb{R}^+$$ such that $$x\not\in A\cup B$$. Consider $$A_x=\{kx+a:k\in\mathbb{Q}^+\cup\{0\}\textrm{ and }a\in A\}$$ and $$B_x=\{kx+b:k\in\mathbb{Q}^+\cup\{0\}\textrm{ and }b\in B\}$$. Then $$A\subseteq A_x$$, $$B\subseteq B_x$$ and $$A_x,B_x\subseteq\mathbb{R}^+$$. It is easy to verify that $$A_x$$ and $$B_x$$ are closed under addition and multiplication by a positive rational number.

If either $$A_x\cap B=\emptyset$$ or $$A\cap B_x=\emptyset$$, then $$(A,B)$$ would not be a maximal element of $$(\mathcal{P},\leq)$$. Therefore, $$A_x\cap B\neq\emptyset$$ and $$A\cap B_x\neq\emptyset$$, so there is some $$q_0\in\mathbb{Q}^+$$ and $$a\in A$$ such that $$q_0x+a\in B$$, and there is some $$q_1\in\mathbb{Q}^+$$ and $$b\in B$$ such that $$q_1x+b\in A$$. Note that $$q_0,q_1\neq 0$$ because $$A\cap B=\emptyset$$. It is easy to see that $$q_0q_1x+q_1a+q_0b\in A\cap B$$, so $$A\cap B\neq\emptyset$$, a contradiction. Therefore, $$A\cup B=\mathbb{R}^+$$.