# Subsets of $\mathbb R$ which are closed under addition and multiplication and those which are not and question of existence of bijection

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 $$b$$.

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.

$$M$$ 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.

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