1
大学数学基礎解説
文献あり

Independent Sets and the Continuum Hypothesis

113
0
$$$$

数学論理学アドベントカレンダー2022 の記事を書いてみました。遅れてすみません。

Let $[X]^{<\omega}:=\{Y\subseteq X:|Y|<\aleph_0\}$ denote the collection of all finite subsets of $X$ (also written $\mathscr{P}_\omega(X)$).

Independent set

Let $f:[X]^{<\omega}\to\mathscr{P}(X)$ be any function. A set $A\subseteq X$ is called independent if for all finite $A_0\in[A]^{<\omega}$ and $\alpha\in A\setminus A_0$, we have $\alpha\notin f(A_0)$.

For example, if $\mathbb{F}$ is a field, $\mathbb{X}$ is an $\mathbb{F}$-vector space, and $\operatorname{span}_\mathbb{F}:[\mathbb{X}]^{<\omega}\to\mathscr{P}(X)$ is the $\mathbb{F}$-span, then a set $A\subseteq\mathbb{X}$ is independent with respect to $\operatorname{span}_\mathbb{F}$ iff $A$ is $\mathbb{F}$-linearly independent.

We have a folklore characterization of the Continuum Hypothesis ($\textsf{CH}$).

The following are equivalent:

  1. For any $f:[\mathbb{R}]^{<\omega}\to[\mathbb{R}]^{<\omega}$, there exists an independent set of size $3$.
  2. $\lnot\textsf{CH}$.

In this article, we'll see proofs of the following facts.

  1. For any $k\in\mathbb{N}$ and $f:[\aleph_k]^{<\omega}\to[\aleph_k]^{<\omega}$, there exists an independent set $A\subseteq\aleph_k$ of size $|A|=k+1$.
  2. For $k\in\mathbb{N}$, there exists some $f:[\aleph_k]^{<\omega}\to[\aleph_k]^{<\omega}$ without any independent set $A\subseteq\aleph_k$ of size $|A|=k+2$.
  3. For any Borel $f:[\mathbb{R}]^{<\omega}\to[\mathbb{R}]^{<\omega}$, there exists an independent nonempty perfect set $P\subseteq\mathbb{R}$. (Thus $|P|=2^{\aleph_0}$.)

Items (i) and (ii) imply Theorem 1, since one sees that every $f:[X]^{<\omega}\to[X]^{<\omega}$ admits an independent set of size $3$ iff $|X|\ge\aleph_2$.

Item (iii) illustrates that the above $\textsf{CH}$ phenomenon is orthogonal to descriptive set theory. Indeed, if $f:[\mathbb{R}]^{<\omega}\to[\mathbb{R}]^{<\omega}$ is Borel, then moving to a generic extension with $(|\mathbb{R}|=2^{\aleph_0}>\aleph_\omega)^{V[G]}$ and using item (i), we see from $\mathbf{\Sigma^1_1}$-absoluteness that in $V$, $f$ admits independent sets of size $k$ for any $k\in\mathbb{N}$. I learned about this absoluteness argument from ジタさん's article here .

Proof of (i)

The following proof is from [1, Lemma 2.3].

We proceed via induction on $k$. When $k=0$, let $A=\{\alpha\}$ be any set with $\alpha\notin f(\varnothing)$, and this is an independent set of size $1$.

Let $k>0$ and $f:[\aleph_k]^{<\omega}\to[\aleph_k]^{<\omega}$, and fix $\aleph_{k-1}\subseteq \aleph_k$.

For a set $Y\subseteq\aleph_k$ (later taken to be $Y=\aleph_{k-1}$), we put $f[Y]:=\bigcup_{Y_0\in [Y]^{<\omega}}f(Y_0)$, so that a routine cardinality computation shows that $|f[Y]|\le|Y|+\aleph_0$. We can perform the "closure under $f$" operation
$$\overline{Y}:=Y\cup f[Y]\cup f[f[Y]]\cup\cdots=\bigcup_{n\in\mathbb{N}}f^n[Y]$$
such that $|Y|\le|\overline{Y}|\le|Y|+\aleph_0$ and moreover $f[\overline{Y}]\subseteq\overline{Y}$. This defines the least $\overline{Y}\supseteq Y$ with the property that $f[\overline{Y}]\subseteq\overline{Y}$.

Now since $|\overline{\aleph_{k-1}}|=\aleph_{k-1}<\aleph_k$, we can pick some arbitrary $\alpha\in \aleph_k\setminus \overline{\aleph_{k-1}}$ outside of this closure.

Define the "restriction to $\overline{\aleph_{k-1}}$" function $g:[\overline{\aleph_{k-1}}]^{<\omega}\to [\overline{\aleph_{k-1}}]^{<\omega}$ such that
$$g(X_0)=\big(f(X_0)\cup f(X_0\cup\{\alpha\})\big)\cap \overline{\aleph_{k-1}}\in[\overline{\aleph_{k-1}}]^{<\omega}$$
By induction on $k$, the function $g:[\overline{\aleph_{k-1}}]^{<\omega}\to [\overline{\aleph_{k-1}}]^{<\omega}$ has an independent set $A\subseteq \overline{\aleph_{k-1}}$ of size $k$.

We claim that $A\cup\{\alpha\}$ is an independent set of size $k+1$. It's clear from the $g$-independence of $A$ that if $\beta \in A$ and $A_0\subseteq A\setminus\{\beta\}$, then
$$\beta\notin f(A_0)\cup f(A_0\cup\{\alpha\})$$
Moreover for any $A_0\subseteq A$, we have
$$\alpha\notin f(A_0)\subseteq \overline{\aleph_{k-1}}$$
due to $\overline{\aleph_{k-1}}\supseteq A$ being closed under $f$. So $A\cup\{\alpha\}$ is indeed independent. $\square$

Proof of (ii)

Again we proceed by induction on $k\in\mathbb{N}$.

When $k=0$, we define $f:[\aleph_0]^{<\omega}\to[\aleph_0]^{<\omega}$ via
$$f(X_0)=\begin{cases}\{0,1,\ldots,n-1\}&\text{if }X_0=\{n\}\\\varnothing&\text{else}\end{cases}$$
Then any set $\{m,n\}\subseteq\aleph_0=\mathbb{N}$ of size $2$ cannot be independent since if $m< n$, then $m\in f(\{n\})$.

Let $k>0$. For each $\alpha<\omega_k$, since $|\alpha|\le\aleph_{k-1}$, we can fix by induction a function $f_\alpha:[\alpha]^{<\omega}\to[\alpha]^{<\omega}$ without independent sets of size $k+1$. Then we define $f:[\omega_k]^{<\omega}\to[\omega_k]^{<\omega}$ via
$$f(X_0)=\begin{cases}\varnothing&\text{if }X_0=\varnothing\\f_{\max(X_0)}(X_0\setminus\{\max(X_0)\})&\text{else}\end{cases}$$

For example, when $k=1$, our function $f:[\aleph_1]^{<\omega}\to[\aleph_1]^{<\omega}$ defined above can be made more explicit. For each $\alpha<\omega_1$, fix an enumeration $\alpha=\{\alpha_0,\alpha_1,\ldots\}$ with indices from $\omega$. Then
$$f(X_0)=\begin{cases}\{\alpha_0,\alpha_1,\ldots,\alpha_{n-1}\}&\text{if }X_0=\{\alpha_n<\alpha\}\\\varnothing&\text{else}\end{cases}$$
One may use this explicit form to check that $f$ has no independent set of size $3$. [Hint: Let $\{\alpha_m<\alpha_n<\alpha\}$ be the independent set.]

We claim that $f$ admits no independent set of size $k+2$. Towards a contradiction, assume $X_0\in[\omega_k]^{<\omega}$ is an independent set with $|X_0|=k+2$. Let $\alpha=\max(X_0)$, and so $X_0\setminus\{\alpha\}$ is an independent set for $f_\alpha$ of size $k+1$, which is a contradiction. $\square$

A slight variation of the proofs above gives another equivalent formulation of $\textsf{CH}$:

Let $[X]^{\le\omega}:=\{Y\subseteq X:|Y|\le\aleph_0\}$ denote the collection of all countable subsets of $X$ (also written $\mathscr{P}_{\omega_1}(X)$).

The following are equivalent:

  1. For any $f:[\mathbb{R}]^{<\omega}\to[\mathbb{R}]^{\le\omega}$, there exists an independent set of size $2$.
  2. $\lnot\textsf{CH}$.

Proof of (iii)

Recall that $[\mathbb{R}]^{<\omega}$ is a standard Borel space, when its Borel structure is inherited from the set of representatives $\{(x_i)\in\mathbb{R}^{<\omega}:x_0< x_1<\ldots< x_{n-1}\}$ as a Borel (in fact, open) subset of the Polish space $\mathbb{R}^{<\omega}:=\bigsqcup_{n\in\mathbb{N}}\mathbb{R}^n$.

Fix any Borel function $f:[\mathbb{R}]^{<\omega}\to[\mathbb{R}]^{<\omega}$.

For every positive integer $n>0$, the set
$$R_n:=\{(x_i)\in\mathbb{R}^n:\{x_i\}\in [\mathbb{R}]^{<\omega}\text{ is an independent set of size }n\}$$
is a comeager Borel subset of $\mathbb{R}^n$.

Fix $n>0$, and let
$$C_n:=\{(x_i)\in\mathbb{R}^n:x_0\notin f(\{x_1,\ldots,x_{n-1}\})\}\subseteq\mathbb{R}^n$$
It's clear that $C_n$ is Borel.

Then for each $(x_1,\ldots,x_{n-1})\in\mathbb{R}^{n-1}$, the section
$$C_n^{(x_1,\ldots,x_{n-1})}=\{x_0\in\mathbb{R}:x_0\notin f(\{x_1,\ldots,x_{n-1}\})\}=\mathbb{R}\setminus f(\{x_1,\ldots,x_{n-1}\})$$
is a cofinite subset of $\mathbb{R}$, and so is comeager.

By Kuratowski–Ulam [2, Theorem 8.41], every section of $C_n$ is comeager implies that the set $C_n$ is comeager Borel.

In a similar way and by modifying the proof above, Kuratowski–Ulam also implies that the desired set
$$R_n=\{(x_i)\in\mathbb{R}^n:\forall j\in n,\,\forall x_j\notin A_0\subseteq\{x_i\},\,x_j\notin f(A_0)\cup A_0\}$$
is comeager Borel. Here we may use that comeager Borel sets are closed under finite intersections, and also preimages under canonical projections $\pi:\mathbb{R}^n\twoheadrightarrow\mathbb{R}^m$. $\square$

Now let's recall the definition and some properties of the hyperspace $K(X)$ of (nonempty) compact subsets of $X$ (later taken to be $X=\mathbb{R}$).

The hyperspace $K(X)$

Let $X$ be a Polish space. Denote by $K(X)$ the set of (nonempty) compact subsets of $X$. When $K(X)$ is endowed with the Vietoris topology (see [2, §4.F]), $K(X)$ becomes a Polish space.

More explicitly, if $d$ is a compatible complete metric on $X$, then the Hausdorff metric $d_H$ is a compatible complete metric on $K(X)$ (see [2, Theorem 4.25]).

[2, Exercise 8.8]

Recall that a Polish space $X$ is perfect if $X$ has no isolated points.

If $X$ is a perfect Polish space, then comeagerly many $K\in K(X)$ is (nonempty) perfect.

[2, Theorem 19.1]

For $K\subseteq X$ and $n\in\mathbb{N}$, define $(K)^n:=\{(x_i)\in K^n:\forall i\ne j,\,x_i\ne x_j\}$ to be the set of tuples in $K^n$ with pairwise distinct entries.

Let $X$ be a Polish space. For all $i\in\mathbb{N}$, we fix a natural number $n_i\in\mathbb{N}$ and a comeager subset $R_i\subseteq X^{n_i}$. Then comeagerly many $K\in K(X)$ satisfies that $\forall i\in\mathbb{N},\,(K)^{n_i}\subseteq R_i$.

Fixing the sets $R_n\subseteq\mathbb{R}^n$ from Lemma 4, we can use Lemmas 5 and 6 to find a comeager set of $P\in K(\mathbb{R})$ such that $P\subseteq\mathbb{R}$ is nonempty perfect and $(P)^n\subseteq R_n$ for all $n$. By definition of independent sets and using $(P)^n\subseteq R_n$, we see that any such $P$ is an independent set. $\square$

The same proof above applies to any Universally Baire measurable $f:[\mathbb{R}]^{<\omega}\to\mathbb{R}^{\le\omega}$ as well.

For example, let $\mathbb{F}\subseteq\mathbb{R}$ be a countable subfield of $\mathbb{R}$. The $\mathbb{F}$-span $\operatorname{span}_\mathbb{F}:[\mathbb{R}]^{<\omega}\to\mathbb{R}^{\le\omega}$ can be defined to be a Borel function. So by Example 1, there must exist a nonempty perfect set $P\subseteq\mathbb{R}$ which is $\mathbb{F}$-linearly independent.

Cohen forcing

We can give an explicit combinatorial account of the previous proof of (iii), and some alternative lemmas to replace Lemmas 5 and 6.

For simplicity, instead of using $\mathbb{R}$, we replace it with the Baire space $\omega^\omega$. For our purposes, the Cohen forcing $\mathbb{C}=(\omega^{<\omega},{\prec})$ consists of conditions which are finite segments $p=(x_0,\ldots,x_{n-1})\in\omega^{<\omega}$, and we say that $p\preceq q$ ($p$ is stronger than $q$) if $p$ is an end-extension of $q$.

When $G\subseteq\omega^{<\omega}$ is $\mathbb{C}$-generic, we write the corresponding Cohen real as $x_G:=\bigcup G\in \omega^\omega$.

For a Borel set $B\subseteq\omega^\omega$, we can use $\mathbf{\Sigma^1_1}$-absoluteness to define it in $V[G]$ using the same Borel code, such that $B=B^{V[G]}\cap (\omega^\omega)^V$. Then recall that $B\subseteq\omega^\omega$ is comeager iff $\Vdash x_G\in B^{V[G]}$.

Now we define a notion of forcing that generically adds a nonempty perfect subset $P$ of $\omega^\omega$.

The notion of forcing $\mathbb{P}$ consists of conditions $p$ such that for some $n\in\mathbb{N}$, $p$ is a function $p:2^{< n}\to\omega^{<\omega}$, satisfying for all $s,t\in 2^{< n}$,
$$s\subseteq t\;\Rightarrow\;p(s)\subseteq p(t)$$
where $\subseteq$ denotes end-extension.

For conditions $p,q\in\mathbb{P}$, we say that $p\preceq q$ ($p$ is stronger than $q$) if $p$ is an end-extension of $q$, that is $q=p\upharpoonright\operatorname{dom}(q)$.

When $G\subseteq\mathbb{P}$ is generic, we can define a function $\varphi_G:=\lim G:2^\omega\to \omega^\omega$ such that for all $x\in (2^\omega)^{V[G]}$, $\varphi_G(x):=\bigcup_{n\in\mathbb{N}}(\bigcup G)(x\upharpoonright n)\in\omega^\omega$. (We will show $\operatorname{dom}(\varphi_G(x))=\omega$ in Lemma 7.)

This $\varphi_G$ is clearly continuous, and we will see that in fact it is an embedding, so that $P_G:=\operatorname{image}(\varphi_G)\subseteq\omega^\omega$ is a nonempty perfect set. The main lemma is the following:

For all $x_0,x_1,\ldots,x_{n-1}\in (2^\omega)^{V[G]}$ pairwise distinct ($i\ne j\Rightarrow x_i\ne x_j$), $(\varphi_G (x_0),\varphi_G(x_1),\ldots,\varphi_G (x_{n-1}))$ is $\mathbb{C}^n$-generic over $V$.

The proof is mostly standard, and the only nontrivial step is the following.

Let $p\in\mathbb{P}$ with $\operatorname{dom}(p)=2^{< n}$, and let $D\subseteq\mathbb{C}^{2^n}$ be a dense set. Then there exists some $q\prec p$ with $\operatorname{dom}(q)=2^{\le n}$ and $\langle q(t):t\in 2^n\rangle\in D$.

Lemma 8

Consider $\hat{p}:=\langle p(t\upharpoonright_{(n-1)}): t\in 2^n\rangle\in\mathbb{C}^{2^n}$. Then since $D\subseteq\mathbb{C}^{2^n}$ is dense, there exists some $\hat{q}\preceq\hat{p}$ such that $\hat{q}\in D$. Define $q:2^{\le n}\to \omega^{<\omega}$ such that
$$q(t)=\begin{cases}p(t)&\text{if }t\in 2^{< n}\\\hat{q}(t)&\text{if }t\in 2^n\end{cases}$$
Now we finish by checking that $q\in \mathbb{P}$, $q\prec p$. $\square$

Lemma 7

Fix $m\in\mathbb{N}$ such that $x_i\upharpoonright m\in 2^m$ are pairwise distinct for $i\in n$.

For any open dense set $D\subseteq\mathbb{C}^n$ and $k\in\mathbb{N}$, define $\hat{D}_k\subseteq\mathbb{C}^{k}$ such that
$$(p_0,\ldots,p_{k-1})\in \hat{D}_k\;\Leftrightarrow\; \forall i_0\ne \cdots \ne i_{n-1}\in k,\, (p_{i_0},\ldots,p_{i_{n-1}})\in D$$
Note that for every $i_0\ne\cdots\ne i_{n-1}\in k$ (pairwise distinct), the set $\{(p_i)\in \mathbb{C}^k:(p_{i_0},\ldots,p_{i_{n-1}})\in D\}$ is open dense in $\mathbb{C}^k$, and so $\hat{D}_k$ as a finite intersection of these $\binom{k}{n}\cdot n!$ many open dense sets is open dense in $\mathbb{C}^k$.

Define $E\subseteq\mathbb{P}$ such that for any $p\in\mathbb{P}$ with $\operatorname{dom}(p)=2^{\le k}$,
$$p\in E\;\Leftrightarrow\;\left(k\ge m\right)\land\left( \langle p(t):t\in 2^k\rangle\in \hat{D}_{2^k}\right)$$
Since we've shown that $\hat{D}_{2^k}$ is open dense in $\mathbb{C}^{2^k}$, Lemma 8 implies that $E\subseteq\mathbb{P}$ is dense, so we can fix some $p\in E\cap G$ and let $\operatorname{dom}(p)=2^{\le k}$.

Since $k\ge m$, $x_i\upharpoonright k\in 2^k$ are pairwise distinct for $i\in n$. By definition of $E$ and $\hat{D}_{2^k}$, we have
$$\bigl(p(x_0\upharpoonright k),p(x_1\upharpoonright k),\ldots,p(x_{n-1}\upharpoonright k)\bigr)\in D$$
Since $p\in G$, we also have $(p(x_i\upharpoonright k))\subseteq (\varphi_G(x_i))$ as an initial segment, so $(\varphi_G(x_i))$ meets $D$ as desired. $\square$

When combined with Lemma 4, Lemma 7 implies that $P_G\subseteq(\omega^\omega)^{V[G]}$ is an independent nonempty perfect set. Then the statement “there exists an independent nonempty perfect set” is $\mathbf{\Sigma^1_2}$, so $\mathbf{\Sigma^1_2}$-absoluteness implies that it holds in $V$ as well. This gives an alternative proof of (iii). $\square$

Infinite independent sets

Questions of the form “Which $f:[\kappa]^{<\omega}\to[\kappa]^{<\omega}$ admits an infinite independent set?” is once again less related to the question of $(\textsf{G})\textsf{CH}$, but more related to partition properties of cardinals. For example:

[3, Theorem 45.2]

Let $\kappa$ be a Ramsey cardinal and let $\lambda <\kappa$. Then every $f:[\kappa]^{<\omega}\to [\kappa]^{<\lambda}$ admits an independent set of size $\kappa$.

Compared with the main phenomema (i) and (ii), the consistency of “every $f:[\aleph_\omega]^{<\omega}\to [\aleph_\omega]^{<\omega}$ admits an infinite independent set” seems to be still unclear. The best result I could find is the following:

[4, Theorem 2]

Assume $\textsf{V}=\textsf{L}$. The following are equivalent:

  1. Every $f:[\kappa]^{<\omega}\to[\kappa]^{\le\omega}$ admits an independent set of size $\omega$.
  2. $\kappa\rightarrow(\omega)^{<\omega}_{2}$.

So if $\textsf{V}=\textsf{L}$, then $\kappa=\aleph_\omega$ does not satisfy the critetion of this last theorem, see eg. [5, Proposition 7.15].

Strong Fubini's Theorem

We conclude this article with a connection to the strong Fubini's theorem. I first learned about strong Fubini's theorem from でぃぐさん's article [6].

Let $\textsf{null}\subseteq\mathscr{P}(\mathbb{R})$ denote the set of Lebesgue null sets in $\mathbb{R}$.

The following are equivalent:

  1. Every $f:[\mathbb{R}]^{<\omega}\to\textsf{null}$ admits an independent set of size $2$.
  2. For any function $f:[0,1]^2\to[0,1]$, it holds that
    $$\int_0^1\!\int_0^1 f\:dx\,dy=\int_0^1\!\int_0^1 f\:dy\,dx$$
    whenever both integrals exist in the sense of Lebesgue integration.

The reader may compare this statement with the statement of Theorem 3, where every $f:[\mathbb{R}]^{<\omega}\to[\mathbb{R}]^{\le\omega}$ admitting an independent set of size $2$ is equivalent to $\lnot\textsf{CH}$.

It's easy to see that statement (1) from Proposition 11 is equivalent to the following statement:

(1') Every $\hat{f}:[0,1]\to\textsf{null}$ admits an independent pair $\alpha,\beta\in [0,1]$, in the sense that $\alpha\notin \hat{f}(\beta)$ and $\beta\notin \hat{f}(\alpha)$.

  • Here, (1) ⇒ (1') is easy since given any $\hat{f}:[0,1]\to\textsf{null}$, we can let $f:[\mathbb{R}]^{<\omega}\to\textsf{null}$ be defined via $f(\{\cot\pi\alpha\})=\cot(\pi\cdot\hat{f}(\alpha))$ if $\alpha\in(0,1)$ and $f(S)=\varnothing$ otherwise, and so any $f$-independent set $\{\cot\pi\alpha,\cot\pi\beta\}\subseteq\mathbb{R}$ of size $2$ gives an $\hat{f}$-independent pair $\alpha,\beta\in [0,1]$.
  • Next, (1') ⇒ (1) is also easy since if $f:[\mathbb{R}]^{<\omega}\to\textsf{null}$, then letting for every $\alpha\in[0,1]$, $\hat{f}(\alpha):=\{\alpha\}\cup f(\varnothing)\cup f(\{\alpha\})$, we see that any $\hat{f}$-independent pair $\alpha,\beta\in[0,1]$ gives an $f$-independent set $\{\alpha,\beta\}$ of size $2$.

A proof from [6, Lemma 2] showed that statement (2) of Proposition 11 is equivalent to the following statement:

(2') There does not exist a set $\hat{D}\subseteq [0,1]^2$, such that:

  • For every $x\in [0,1]$, the vertical section $\hat{D}_x:=\{y\in[0,1]:(x,y)\in \hat{D}\}$ is Lebesgue conull;
  • For every $y\in [0,1]$, the horizontal section $\hat{D}^y:=\{x\in [0,1]:(x,y)\in \hat{D}\}$ is Lebesgue null.

Note that in [6, Lemma 2], the quantifiers for the sections' coordinates $x,y$ are "for almost every", but we can always modify $D$ over a Lebesgue null subset of $[0,1]^2$ to get to a set $\hat{D}$ of our desired form, where we can quantify over every vertical/horizontal section.

It remains to show that (1') ↔ (2'). The following proofs are essentially contained in [6, Theorem 3].

$\lnot(1')\Rightarrow\lnot(2')$

Let $f:[0,1]\to\textsf{null}$ admit no independent pair. That is, for all $\alpha,\beta\in [0,1]$, we have $\alpha\in f(\beta)$ or $\beta\in f(\alpha)$.

Define the set $D:=\{(x,y)\in[0,1]^2:x\in f(y)\}$.

  • Fix any $x\in [0,1]$. For any $y\in [0,1]$, we have either $x\in f(y)$ so then $y\in D_x$, or $y\in f(x)$. So every possible $y\in[0,1]$ is covered by $D_x\cup f(x)=[0,1]$. Then since $f(x)\in\textsf{null}$, the section $D_x\supseteq [0,1]\setminus f(x)$ is Lebesgue conull.
  • Fix any $y\in [0,1]$. Then the section $D^y=f(y)\in\textsf{null}$ is Lebesgue null.

So $D\subseteq[0,1]^2$ violates (2'). $\square$

$\lnot(2')\Rightarrow\lnot(1')$

Let $D\subseteq[0,1]^2$ violate (2'). So for every $x,y\in[0,1]$, we have $[0,1]\setminus D_x\in\textsf{null}$ and $D^y\in\textsf{null}$.

Define the function $f:[0,1]\to\textsf{null}$ such that
$$f(x):=([0,1]\setminus D_x)\cup D^x$$

Fix any $\alpha,\beta\in[0,1]$.

  • If $(\alpha,\beta)\in D$, then $\alpha\in D^\beta\subseteq f(\beta)$.
  • If $(\alpha,\beta)\notin D$, then $\beta\in [0,1]\setminus D_\alpha\subseteq f(\alpha)$.

So $f$ admits no independent pair. $\square$

参考文献

[1]
John T. Baldwin, Sy D. Friedman, Martin Koerwien, Michael C. Laskowski, Three Red Herrings around Vaught’s Conjecture, Transactions of the American Mathematical Society, 2016, pp. 3673-3694
[2]
Alexander S. Kechris, Classical Descriptive Set Theory, Graduate Texts in Mathematics, Springer-Verlag, 1995
[3]
P. Erdős, A. Hajnal, A. Máté, R. Rado, Combinatorial Set Theory: Partition Relations for Cardinals, Studies in Logic and the Foundations of Mathematics, North-Holland, 1984
[4]
K.J. Devlin, J.B. Paris, More on the free subset problem, Annals of Mathematical Logic, 1973, pp. 327-336
[5]
Akihiro Kanamori, The Higher Infinite: Large Cardinals in Set Theory from their Beginnings, Springer Monographs in Mathematics, Springer, 2009
投稿日:20221222

この記事を高評価した人

高評価したユーザはいません

この記事に送られたバッジ

バッジはありません。

投稿者

duyaa
duyaa
6
228
中国人です。今はAlexander Kechris先生の弟子として、カリフォルニア工科大学で博士課程を受けています。

コメント

他の人のコメント

コメントはありません。
読み込み中...
読み込み中