Is that possible to partition $(\Bbb R,+)$ into 4 additively closed subsets? Is that possible to partition $(\Bbb R,+)$ into 4 additively closed subsets?
It is easy to partition it to 1 and 2 and 3 additively closed subsets. Fore example for 2 subsets we have: $(\Bbb R^{\geq 0},+)\cup(\Bbb R^{< 0},+)$. But Is that possible for 4 additively closed subsets? What about $k$ additively closed subsets?
 A: For completeness I’ll add an answer showing that in fact $\Bbb R$ can be partitioned into $\kappa$ sets closed under addition for any non-zero cardinal $\kappa\le 2^\omega=\mathfrak{c}$. (Of course this uses the axiom of choice.) Start with a Hamel basis $B=\{b_\xi:\xi<2^\omega\}$ for $\Bbb R$ over $\Bbb Q$. For each $x\in\Bbb R\setminus\{0\}$ there is a unique finite $B_x\subseteq B$ such that $x$ is a linear combination with non-zero rational coefficients of the members of $B_x$; let $B_x^+$ be the set of members of $B_x$ whose coefficients in that linear combination are positive. For each $\eta<2^\omega$ let
$$A_\eta=\big\{x\in\Bbb R:\min\{\xi<2^\omega:b_\xi\in B_x\}=\eta\text{ and }b_\eta\in B_x^+\big\}\,;$$
$b_\eta\in A_\eta$, so $A_\eta\ne\varnothing$, and $A_\eta$ is clearly closed under addition.
Now let $\kappa\le 2^\omega$ be a cardinal, and let $$D=\Bbb R\setminus\bigcup_{\xi<\kappa}A_\xi\,.$$
Clearly $\{D\}\cup\{A_\xi:\xi<\kappa\}$ is a partition of $\Bbb R$ into $\kappa$ parts if $\kappa\ge\omega$, and into $\kappa+1$ parts if $\kappa<\omega$, and it only remains to show that $D$ is closed under addition. But $x\in D$ iff either

*

*$\min\{\xi<2^\omega:b_\xi\in B_x\}\ge\kappa$, or

*$\eta=\min\{\xi<2^\omega:b_\xi\in B_x\}<\kappa$ and $b_\eta\notin B_x^+$, or

*$x=0$,

and it’s easy to check that the set of real numbers satisfying one of these conditions is closed under addition.
A: You can partition $(\Bbb R^n,+)$ into the three subsemigroups $H^n_+=\{x\in\Bbb R^n\,:\, x_n>0\}$, $H^n_-=\{x\in\Bbb R^n\,:\,x_n<0\}$ and $\Bbb R^{n-1}\times\{0\}$. Therefore, if $\Bbb R^{n-1}$ can be partitioned into $k$ subsemigroups, then $\Bbb R^n$ can be partitioned into $k+2$ subsemigroups. Now, let $m$ be the smallest positive natural number such that $(\Bbb R,+)$ cannot be partitioned in $m$ subsemigroups. As you said, $m\ge3$, and by definition $(\Bbb R,+)$ can be partitioned in $m-2$ subsemigroups. But then $(\Bbb R^2,+)$ can be partitioned in $m$ subsemigroups. But thanks to AC (axiom of choice), $(\Bbb R,+)$ and $(\Bbb R^2,+)$ are isomorphic as groups, and therefore $(\Bbb R,+)$ can be partitioned into $m$ subsemgroups as well, against the hypothesis on $m$. Therefore $(\Bbb R,+)$ can be partitioned into any finite number of subsemigroups.
