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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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  • $\begingroup$ 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. $\endgroup$
    – Omar
    Nov 25, 2012 at 19:18

4 Answers 4

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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 $i<j$ 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 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 $X\cup Y=\mathbb R^+$.

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    $\begingroup$ 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 $x+y\in X\cap Y$). $\endgroup$
    – Asaf Karagila
    Nov 25, 2012 at 19:38
  • $\begingroup$ 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 $x+u$? $\endgroup$
    – Mirko
    May 17, 2017 at 23:57
  • $\begingroup$ 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. $\endgroup$
    – Mirko
    May 18, 2017 at 3:04
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    $\begingroup$ @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 $X\cap Y$. $\endgroup$ May 18, 2017 at 3:28
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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.

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

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    $\begingroup$ That doesn't answer the question of using Zorn's lemma specifically. $\endgroup$ Jul 10, 2018 at 10:33
  • $\begingroup$ @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. $\endgroup$
    – Ur Ya'ar
    Jul 12, 2018 at 8:38
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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, $A\cup B=\mathbb{R}^+$.

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