11
$\begingroup$

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

$\endgroup$
7
  • 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

7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.