# Building the real from Dedekind finite sets

It is well known that the real numbers can be countable union of countable sets by starting with GCH and taking a finite support permutations while collapsing all of $$\aleph_n$$ for natural $$n$$.

The idea relies heavily on the fact that all reals in the symmetric model "come from" the $$\aleph_n$$ of the ground, which are always well orderable.

It is also known that the reals may contain a Dedekind finite set, so they are Dedekind finite union of disjoint sets (if $$\mathfrak p<\mathfrak c$$ is Dedekind finite then $$\frak p\times c=c$$ so we may partition $$\Bbb R$$ into $$\frak p$$ many parts)

Is it possible to get a "better" construction of the reals using Dedekind finite sets? In particular can we have that the reals are:

• Countable union of Dedekind finite sets

• Dedekind finite union of sets smaller than the continuum > this appears to be true in Cohen's model (thanks Wojowu and Asaf)

• Dedekind finite union of Dedekind finite sets

• Dedekind finite union of countable sets

• The answer is yes for point 2: continuum can be union of two sets of cardinality smaller than continuum Mar 26 at 15:05
• To add to what @Wojowu said, this happens in the Cohen model. Mar 26 at 18:34
• An old question of mine is related: mathoverflow.net/questions/157204/… Mar 26 at 19:02
• The answer to 1 is no, because there are no Dedekind-finite-to-one functions from $\mathscr{P}(\omega)$ to $\omega$; see Theorem 3.5 of my paper Generalizations of Cantor's Theorem in ZF. Mar 26 at 19:12
• If there is a Dedekind finite subset of $\mathbb{R}$ which maps onto $\mathbb{R}$, then $\mathbb{R}$ will be a Dedekind finite union of pairs, and then the answers to 3 and 4 will be yes. But I do not know whether there can be such a subset. Mar 26 at 19:19

Q1: There is no such partition. Let $$\langle X_n \rangle$$ be a countable partition of $$\mathbb{R}.$$ We will construct $$n,$$ an open interval $$I,$$ and an injection $$g: \omega \rightarrow I \cap X_n$$ with $$\text{rng}(g)$$ dense in $$I.$$ We recursively construct $$\langle (p_{\alpha}, S_{\alpha}): \alpha<\omega_1 \rangle,$$ with $$p_{\alpha} \in \mathbb{R}$$ and $$S_{\alpha}=\{p_{\xi}: \xi<\alpha\}.$$ At stage $$\alpha,$$ if for some $$n,$$ $$X_n \cap S_{\alpha}$$ is somewhere dense, set $$p_{\alpha}=0.$$ Otherwise, follow the procedure of the proof of the Baire category theorem (with respect to a basis $$\langle U_k \rangle$$ of rational open intervals) to construct $$p_{\alpha} \in \mathbb{R} \setminus \left ( 0 \cup \bigcup_{n<\omega} \overline{X_n \cap S_{\alpha}} \right ).$$

Consider the limit ordinal $$\beta = \{\alpha<\omega_1: p_{\alpha} \neq 0\}.$$ Define a surjective partial map $$h: \omega^2 \rightharpoonup \beta$$ by setting $$h(k, n) = \alpha$$ if $$\alpha$$ is least such that $$p_{\alpha} \in U_k \cap X_n.$$ In particular, $$\beta$$ is countable, and least such that $$p_{\beta}=0.$$ Let $$(n, k)$$ be lexicographically least such that $$X_n \cap S_{\beta}$$ is dense in $$U_k.$$ Then set $$I=U_k$$ and use $$h$$ to construct bijective $$g: \omega \rightarrow X_n \cap S_{\beta} \cap I.$$

Q2: Indeed, in the Cohen model there is a partition of $$\mathbb{R}$$ into two sets of strictly smaller cardinality. The Bernstein set $$B$$ in Theorem 1.7 here works, by a minor adjustment of their argument. Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be injective, and $$b \in [A]^{<\omega}$$ the minimal parameter from which $$f$$ is constructible. Then carry out their argument in terms of $$f$$ and $$b$$ rather than $$T$$ and $$a$$ to show that $$B \cap \text{rng}(f)$$ and $$B \setminus \text{rng}(f)$$ are nonempty.

Q3&4: Consistently, $$\mathbb{R}$$ is a Dedekind finite union of pairs. Start in $$L,$$ let $$G=\langle c_{\alpha}: \alpha<\omega_1 \rangle$$ be an $$L$$-generic sequence of Cohen reals. In $$L[G],$$ let $$A=\{c_{\alpha}: \alpha<\omega_1\}$$ and $$R=\bigcup_{a \in [A]^{<\omega}} \mathbb{R}^{L[a]}.$$

Let $$\pi=(\pi_1, \pi_2) \in L$$ be the standard bijection from $$\omega_1 \rightarrow \omega_1^2.$$ We partition $$A$$ by letting $$A_{\alpha}=\{c_{\xi}: \xi \in \pi_1^{-1}(\alpha)\}.$$ Let $$M=L(R, \langle A_{\alpha}: \alpha<\omega_1\rangle).$$ We will show $$M$$ has our desired property.

Claim: $$R=\mathbb{R}^M.$$

Proof of claim: Fix $$r \in \mathbb{R}^M.$$ There is $$\varphi,$$ an ordinal $$\gamma,$$ and $$a=\{c_{\alpha_0}< \ldots such that, letting $$G'=G \restriction \omega_1 \setminus \{\alpha_j\},$$ we have for all $$n<\omega$$ that

\begin{align*}n \in r &\Leftrightarrow M \models \varphi(n, \gamma, c_{\alpha_0}, \ldots, c_{\alpha_i}, \langle A_{\alpha} \rangle) \\&\Leftrightarrow L[G] =L[c_{\alpha_0}, \ldots, c_{\alpha_i}][G'] \models \varphi^{L(R, \langle A_{\alpha} \rangle)}(n, \gamma, c_{\alpha_0}, \ldots, c_{\alpha_i}, \langle A_{\alpha} \rangle) \\&\Leftrightarrow L[c_{\alpha_0}, \ldots, c_{\alpha_i}] \models \text{Add}(\omega, \omega_1 \setminus \{\alpha_j\}) \Vdash \varphi^{L(R, \langle A_{\alpha} \rangle)}(n, \gamma, c_{\alpha_0}, \ldots, c_{\alpha_i}, \langle A_{\alpha} \rangle), \end{align*} the last forward implication justified by a standard Cohen homogeneity argument, using the invariance of $$R$$ and $$\langle A_{\alpha} \rangle$$ under $$\pi_1$$-preserving permutations of $$\omega_1.$$

Thus, $$r \in \mathbb{R}^{L[c_{\alpha_0}, \ldots, c_{\alpha_i}]} \subset R.$$ This proves the claim. $$\square$$

Suppose some $$r \in R$$ codes an injective sequence $$\langle c_n: n<\omega \rangle \subset A.$$ Then $$\{c_n\} \subset L[a]$$ for some $$a \in [A]^{<\omega},$$ contradicting the mutual genericity of the Cohen reals.

Thus, $$A$$ is Dedekind finite in $$M,$$ which implies $$[A]^{<\omega} \subset [\mathbb{R}]^{<\omega} \equiv \mathbb{R}$$ is also Dedekind finite. We define a surjection $$f: [A]^{<\omega} \setminus \{\emptyset\} \rightarrow R$$ by sending $$a$$ with $$\max a \in A_{\alpha}$$ to the $$\alpha\text{th}$$ real in $$L[a \setminus \{\max a\}].$$ Then

$$R=\bigcup_{a \in \text{dom} f \subset \mathbb{R}} \{a, f(a)\chi_{f^{-1}(\mathbb{R} \setminus \text{dom} f)}(a)\}$$ is as desired.

• Thanks for the answer! I still didn't have time to go through the answer in depth, but about the last paragraph, $A^{<\omega}$ is not Dedekind finite (as witness by e.g. constant $a$ sequence of length $n$ for fixed $a$). I think we want to look only at $A^{1-1}×A$, the surjectivity argument should still follow
– Holo
Mar 29 at 11:37
• You’re right. I’ve rephrased it in terms of $[A]^{<\omega}.$ Mar 29 at 13:44