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Suppose that $M$ is the set of all subsets of $\mathbb R$ which are closed under ordinary addition and multiplication as defined over $\mathbb R$ and suppose that $N$ is the set of all subsets of $\mathbb R$ which are either not closed under addition or multiplication (or both).

Is there a bijection $b: M \to N$?

I think that the only $n$-element set in $M$ is $\{0\}$ and that all the other sets in $M$ are infinite but there is also much of infinite subsets of $\mathbb R$ in $N$ so $N$ consists of almost all finite subsets of $\mathbb R$ and of much of infinite ones so if I had to guess I would say that there is no such $b$.

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As you know from your previous question, the set of subsets of $\Bbb{R}$ that are not both closed under addition and scalar multiplication are equinumerous with the set of subsets of $\Bbb{R}$. So, the question is, is the set of subsets of $\Bbb{R}$ that are both closed under $+$ and $\times$ also equinumerous with the elements of $2^{\Bbb R}$?

To answer this, I'm going to use mathematics that is a little out of my depth. "Recall" that the transcendence degree of the field $\Bbb{R}$ over $\Bbb{Q}$ is the cardinality of the continuum, meaning that there exists an algebraically independent subset $A \subseteq \Bbb{R}$ such that $|A| = |\Bbb{R}|$. In particular, this implies that no non-trivial combination of sums and products of elements of $A$ will ever produce an element of $A$.

Now, from any subset $B \subseteq A$, we can consider $B'$, the closure of $B$ under both sums and products. I claim that the map $B \mapsto B'$ is an injective map from $2^A$ into the set of subsets of $\Bbb{R}$ that are closed under addition and scalar multiplication.

Suppose $B, C \subseteq A$, $B' = C'$, and $x \in B$. Then $x \in B'$, and hence $x \in C'$. Thus, there must exist a polynomial $p : \Bbb{R}^n \to \Bbb{R}$ with rational (in fact, integer) coefficients, and $x_1, \ldots, x_n \in C$ such that $x = p(x_1, \ldots, x_n)$. If $x$ is not at least equal to one of the $x_1, \ldots, x_n \in C$, then we have a polynomial expression of one element of $A$ ($x$) that can be expressed as a polynomial expression of other elements of $A$ ($x_1, \ldots, x_n$), contradicting algebraic independence.

Note that $|2^A| = |2^\Bbb{R}|$, and we have just shown that we can inject this set into the set of sets in $2^{\Bbb{R}}$ that are closed under addition and multiplication, which is, of course, included in $2^{\Bbb{R}}$. Thus, these sets are equinumerous to the set of all subsets of $\Bbb{R}$, and hence your two sets should (in theory) have a bijection between them.

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$M$ and $N$ have the same cardinality as the powerset $\Bbb{P}(\Bbb{R})$, so the bijection $b : M \to N$ does exist.

By the Schröder–Bernstein theorem, to prove what I have just claimed I have to exhibit injections $g : \Bbb{P}(\Bbb{R}) \to M$ and $h: \Bbb{P}(\Bbb{R}) \to N$. To do this, let $\{ t_x \mid x \in \Bbb{R}\}$ be an $\Bbb{R}$-indexed family of elements of $\Bbb{R}$ that are algebraically independent over $\Bbb{Q}$. For any proper subset $X$ of $\Bbb{R}$ put $g(X) = \Bbb{R}[t_x \mid x \in X]$, the subring of $\Bbb{R}$ generated by the $t_x$ with $x \in X$ and let $h(X) = g(X) \cup \{s_X\}$ where $s_X$ is some member of $\Bbb{R} \setminus g(X)$. For $X = \Bbb{R}$, put $g(X) = \{0\}$ and $h(X) = \{1\}$. Then $g$ and $h$ provide the required injections of $\Bbb{P}(\Bbb{R})$ into $M$ and $N$ respectively.

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  • $\begingroup$ What does it mean algebraically independent elements? $\endgroup$ – user750262 Feb 14 at 1:25
  • $\begingroup$ See en.wikipedia.org/wiki/Algebraic_independence. (Here I mean algebraically independent over $\Bbb{Q}$ as is usual when talking about $\Bbb{R}$ or $\Bbb{C}$.) $\endgroup$ – Rob Arthan Feb 14 at 1:26

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