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

  • 4
    $\begingroup$ The answer is yes for point 2: continuum can be union of two sets of cardinality smaller than continuum $\endgroup$
    – Wojowu
    Mar 26 at 15:05
  • 1
    $\begingroup$ To add to what @Wojowu said, this happens in the Cohen model. $\endgroup$
    – Asaf Karagila
    Mar 26 at 18:34
  • $\begingroup$ An old question of mine is related: mathoverflow.net/questions/157204/… $\endgroup$ Mar 26 at 19:02
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 26 at 19:12
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 26 at 19:19

1 Answer 1


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 $a=\{c_{\alpha_0}< \ldots<c_{\alpha_i}\}\in [A]^{<\omega}$ 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.

  • $\begingroup$ 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 $\endgroup$
    – Holo
    Mar 29 at 11:37
  • $\begingroup$ You’re right. I’ve rephrased it in terms of $[A]^{<\omega}.$ $\endgroup$ Mar 29 at 13:44

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