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  1. I know that rational numbers are order-isomorphic to real algebraic numbers. Does it imply that irrational numbers are order-isomorphic to real transcendental numbers?
  2. I know that the order type of rationals $\eta$ is a homogeneous order type (meaning that for any two its elements there is an automorphism that sends one to another). Are the order types of irrationals and real transcendental numbers homogeneous as well?
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  • $\begingroup$ For both the rationals and the real algebraics, there is a way of viewing inclusion into $\mathbb{R}$ as giving a "Dedekind-completion". Let $\varphi:\mathbb{Q} \to \mathbb{A}$ (where $\mathbb{A}$ denotes the real algebraics) be an order isomorphism. A suitable uniqueness result for Dedekind completions then implies there is a unique order isomorphism $\varphi^* \ : \mathbb{R} \to \mathbb{R}$ extending $\varphi$. This map $\varphi^*$ then restricts to an order isomorphism between $\mathbb{R} \setminus \mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{A}$. $\endgroup$
    – Mike F
    Feb 7, 2012 at 2:10

2 Answers 2

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Here’s an answer to part of (2).

The order type of $\mathbb{P}$, the irrationals, is homogeneous, because $\langle\mathbb{P},\le\rangle$ is order-isomorphic to $\langle\mathbb{Z}^\omega,\preceq\rangle$, where $\preceq$ is the lexicographic order, which is order-homogeneous.

To construct the order-isomorphism, enumerate the rationals as $\mathbb{Q}=\{q_n:n\in\omega\}$. Recursively construct open intervals $I_\sigma$ for $\sigma\in\mathbb{Z}^{<\omega}$ to satisfy the following conditions.

  1. $I_{\langle\rangle}=\mathbb{R}$.
  2. If $\langle\rangle\ne\sigma\in\mathbb{Z}^{<\omega}$, $I_\sigma$ is a non-empty open interval with rational endpoints.
  3. For each $n\in\omega$, $q_n$ is an endpoint of some $I_\sigma$ with $|\sigma|\le n+1$.
  4. For each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, $I_{\sigma^\frown n}\subseteq I_\sigma$.
  5. For each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, the right endpoint of $I_{\sigma^\frown n}$ is the left endpoint of $I_{\sigma^\frown (n+1)}$.
  6. For each $\sigma\in\mathbb{Z}^{<\omega}$, $\{I_{\sigma^\frown n}:n\in\mathbb{Z}\}$ covers all of $I_\sigma$ except the endpoints of the intervals $I_{\sigma^\frown n}$.
  7. If $\langle\rangle\ne\sigma\in\mathbb{Z}^{<\omega}$, the length of $I_\sigma$ is less than $2^{-|\sigma|}$.

Clauses (4)-(6) ensure that for each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, $\operatorname{cl}I_{\sigma^\frown n}\subseteq I_\sigma$, so for each $\sigma\in\mathbb{Z}^\omega$, $$\bigcap\limits_{n\in\omega}I_{\sigma\upharpoonright n}\ne\varnothing\;.\tag{1}$$ Clause (7) ensures that the intersection in $(1)$ is at most a singleton, so for each $\sigma\in\mathbb{Z}^\omega$ there is a unique $h(\sigma)\in\mathbb{R}$ such that $$\bigcap\limits_{n\in\omega}I_{\sigma\upharpoonright n}=\{h(\sigma)\}\;.$$ Finally, clause (3) ensures that $h(\sigma)\in\mathbb{P}$, so $h:\mathbb{Z}^\omega\to\mathbb{P}$.

To see that $h$ is an injection, suppose that $\sigma,\tau\in\mathbb{Z}^\omega$, and $\sigma\ne\tau$. Let $n\in\omega$ be minimal such that $\sigma\upharpoonright n\ne\tau\upharpoonright n$; then by construction $I_{\sigma\upharpoonright n}\cap I_{\tau\upharpoonright n}=\varnothing$, so $h(\sigma)\ne h(\tau)$. To see that $h$ is surjective, simply observe that for any $x\in\mathbb{P}$ and any $n\in\omega$ there is a unique $\sigma\in\mathbb{Z}^n$ such that $x\in I_\sigma$. Thus, $h$ is a bijection, and it only remains to check that $h$ is order-preserving.

Suppose that $\sigma,\tau\in\mathbb{Z}^\omega$ with $\sigma\prec\tau$. Let $n\in\omega$ be minimal such that $\sigma(n)\ne\tau(n)$, and let $\varphi=\sigma\upharpoonright n=\tau\upharpoonright n$. Then $h(\sigma),h(\tau)\in I_\varphi$, $h(\sigma)\in I_{\varphi^\frown \sigma(n)}$, $h(\tau)\in I_{\varphi^\frown \tau(n)}$, and $\sigma(n)<\tau(n)$, so $h(\sigma)<h(\tau)$ by clause (5).

Finally, to see that $\langle\mathbb{Z}^\omega,\preceq\rangle$ is order-homogeneous, let $\sigma,\tau\in\mathbb{Z}^\omega$, and define $\delta\in\mathbb{Z}^\omega$ by $\delta(n)=\tau(n)-\sigma(n)$. Then the shift $$s:\mathbb{Z}^\omega\to\mathbb{Z}^\omega:\varphi\mapsto\langle\varphi(n)+\delta(n):n\in\omega\rangle$$ is an order-automorphism of $\langle\mathbb{Z}^\omega,\preceq\rangle$ taking $\sigma$ to $\tau$.

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  • $\begingroup$ The order isomorphisms of the set of rationals may be mere translations. That doesn't work with the irrationals. So a question is whether they can be made to mesh in some other elegant way with ordinary arithmetic operations. $\endgroup$ Feb 9, 2012 at 16:44
  • $\begingroup$ @Michael: I thought about that a little when I first saw the question, but I didn’t come up with any good ideas. $\endgroup$ Feb 9, 2012 at 18:56
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Suppose $f$ is a strictly increasing bijection function from the linearly ordered set of all rational numbers to the linearly ordered set of all real algebraic numbers. Let $g:\mathbb R \to \mathbb R$ be defined as follows:

$$ g(x) = \sup \{ f(u) : u \in \mathbb Q \text{ and }u < x \}. $$

If I'm not mistaken, one can show by definition-chasing that the restriction of $g$ to the irrationals is a strictly increasing bijection from that set to the real transcendental numbers.

To see that it's surjective, suppose $y$ is a real transcendental number and let

$$x=\sup\{ f^{-1}(w) : w \in \mathbb Q\text{ and }w < y\}.$$

Since $f$ is a strictly increasing bijection, the set whose sup is taken is a non-empty initial segment of $\mathbb Q$. Since some members of $\mathbb Q$ are $>y$, the complement of the set whose sup is taken is not empty. Hence the set has an upper bound in $\mathbb R$; hence the sup exists. The number $x$ cannot be rational since then $f(x)$ would be a rational number $<y$ and other rationals would be between that number and $y$, and their inverse-images under $f$ would be greater than $x$; but that contradicts the definition of $x$.

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    $\begingroup$ It may be worth mentioning that such an $f$ exists because one can prove via a back and forth argument that "countable, dense in itself, without first or last end points" implies order isomorphic to $\mathbb{Q}$. $\endgroup$ Feb 7, 2012 at 0:05
  • $\begingroup$ I believe that you want to say that ‘their inverse images under $f$ would be greater than $x$’. $\endgroup$ Feb 7, 2012 at 0:35
  • $\begingroup$ @BrianM.Scott : or could I have meant to say that by one argument they're less and by another they're greater, and that's the contradiction? $\endgroup$ Feb 7, 2012 at 0:48
  • $\begingroup$ If so, I recommend changing the wording. As it stands, it really does look like an error for: If $f(x)<y$, and other rationals are between $f(x)$ and $y$, they are greater than $f(x)$, so their inverse images under $f$ are greater than $x$. That contradicts the definition of $f$ as the sup of such inverse images. $\endgroup$ Feb 7, 2012 at 1:10
  • $\begingroup$ OK, I've rephrased it. $\endgroup$ Feb 7, 2012 at 5:13

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