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.
Abstract Algebra – Using Zorn’s Lemma to Show Disjoint Union in $\mathbb R^+$
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Let $x\in\mathbb{R}$. Define $$S_{\leq x}=S~\cap~(-\infty,x]~~~\mathrm{and}~~~S_{>x}=S~\cap~ (x,+\infty)$$ and define $S_{\leq}=\{$ the set of all $x$ such that $S_{\leq x}$ is uncountable $\}$, and $S_>=\{$ the set of all $x$ such that $S_{>x}$ is uncountable $\}$.
$S_{\leq}$ is a non empty$^{(*)}$ interval that extends to infinity on the right, and is therefore of the form $[a,+\infty)$ or $(a,+\infty)$ for some $a\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.
Similarly, $S_>$ is a non empty$^{(*)}$ interval that extends to infinity on the left, and is thus of the form $(-\infty,b]$ or $(-\infty,b)$ for some $b\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.
$^{(*)}$ : I have used some form of choice I believe when I say that $S_>$ and $S_{\leq}$ are non empty, because I use the fact that with some form of choice, the countable union of countable sets is still countable. Is it countable choice? Since $$S=\cup_{n\in\mathbb{N}}S_{\leq n}$$ is uncountable, at least one of the $S_{<n}$ must be uncountable, and so for some natural number $n$ you have $n\in S_{<}\neq\emptyset$.
Let's show that $S_>\cap S_{\leq}$ is non empty. If either one of $S_>$ or $S_{\leq}$ is the whole real line, there is no problem. So let's suppose none of them is the whole of $\mathbb{R}$.
Suppose $x\in S_{\leq}$ that is $S_{\leq x}=S~\cap~(-\infty,x]$ is uncountable. Since $$\bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}\subset S_{\leq x}=S~\cap~(-\infty,x]\subset \{x\}\cup \bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}$$ the already used fact that countable unions of countable sets are countable with some form of choice implies that at least one of the $S_{\leq x -\frac{1}{n}}$ is uncountable, which means that for some $n\in\mathbb{N}^*,~x -\frac{1}{n}\in L$ and thus $S_{\leq}$ is open, that is, $$\mathbf{S_{\leq}=(b,+\infty)}$$ for some real number $b$. Similarly, $$\mathbf{S_{>}=(-\infty,a)}$$ for some real number $a$.
Then $S_>\cap S_{\leq}$ is empty iff $a\leq b$. But this would entail that $$S=S_{\leq a}\cup S_{> a}$$ is countable as the union of two countable sets, since $a \notin (-\infty,a)=S_{\leq}$ means $S_{\leq a}$ is countable, and $a \notin (b,+\infty)=S_{>}$ means $S_{> a}$ countable.
It would help readability of your argument if you said something like this:
$1.$ Let $b$ be in $X$. Then $b-b$ is in $X$. So $0$ is in $X$.
$2.$ Because $0$ is in $X$, for any $b$ in $X$ we have $0-b$ is in $X$. So $-b$ is in $X$.
$3.$ For any $a$ and $b$ in $X$, $a+b=a-(-b)$, so by $(2)$, $a+b$ is in $X$.
Best Answer
First let us recall Zorn's lemma.
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 $X\cup Y=\mathbb R^+$.