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It is well known that $\mathbb{R}^2\ncong \mathbb{R}$. It is also known that $\mathbb{Q}^2\cong \mathbb{Q}$. It is a corollary to Sierpiński's theorem which states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. (A proof can be found here and a discussion here.)

As a consequence from the theorem, for every countable subfield $\mathbb{F}\subset \mathbb{R}$ we know that $\mathbb{F}^2\cong \mathbb{F}\cong \mathbb{Q}$.

My question is: what can be said about the general case? In particular, is $\mathbb{R}$ unique in this sense, i.e.

is it the only ordered field which is not homeomorphic to it's power?

Putting this in another way, I want know if this property is a completeness axiom (in the sense discussed for example here or here).

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    $\begingroup$ It is an easy proposition that every ordered field which is not the reals, is totally disconnected, hence so is it's power. So, topological arguments based on connectednes does not seem to help here... $\endgroup$ – Asaf Shachar Nov 29 '14 at 20:38
  • $\begingroup$ Strictly speaking the question doesn't make sense because of a category error (using "homeomorphic" with ordered fields). Do you want to endow ordered fields with their order topology? And how do you topologize the powers? Do you want the order topology of the product of the orders or do you want the product of topological spaces? Extrapolating from your example I assume that you consider $K^n$ with the $n$-fold product of the order topology on $K$. $\endgroup$ – Johannes Hahn Nov 30 '14 at 15:59
  • $\begingroup$ To condense Johannes Hahn's question: isomorphic in what category? $\endgroup$ – David Roberts Nov 30 '14 at 21:51
  • $\begingroup$ @DavidRoberts: The question does not say “isomorphic”. It says “homeomorphic”, and that’s perfectly unambiguous. $\endgroup$ – Emil Jeřábek supports Monica Nov 30 '14 at 22:31
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    $\begingroup$ Yes, I meant for the fields to be endowed with the order topology, and then use the standard product topology for the power. $\endgroup$ – Asaf Shachar Dec 1 '14 at 9:58
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$\let\ob\overline\let\sset\subseteq\let\nsset\nsubseteq\DeclareMathOperator\dom{dom}$ There are many such fields different from $\mathbb R$.

Proposition: Let $F$ be a topological field of cardinality $2^\kappa$ with a dense subfield $D$ of cardinality $\kappa$. Then there exists an intermediate field $D\sset K\sset F$ of cardinality $2^\kappa$ such that the powers $K^n$ for $n\in\mathbb N$ are pairwise non-homeomorphic.

Interesting choices include:

  • $D=\mathbb Q$, $F\subsetneq\mathbb R$. This will produce an incomplete archimedean ordered field $K$.

  • $D=\mathbb Q(x)$ with $x>\mathbb Q$, $F=\hat D$ (the Scott completion of $D$, which has cardinality $2^\omega$, being a perfect Polish space). This will produce a nonarchimedean OF of cardinality $2^\omega$.

  • Whenever $\kappa^{<\kappa}=\kappa$, there exists an OF $|F|=2^\kappa$ with a dense subfield $|D|=\kappa$. [Under the assumption, there exists a $\kappa$-saturated OF $D$ of cardinality $\kappa$. Then it is easy to construct a nested tree of intervals $\{(a_t,b_t):t\in2^{<\kappa}\}$ such that $b_t-a_t<d_{\mathrm{len}(t)}$, where $\lim_{\alpha\to\kappa}d_\alpha=0$. Any path through the tree defines a Cauchy net, aka good cut, thus the completion $F=\hat D$ has cardinality $2^\kappa$.]

Proof: The main point is that a continuous function $K^n\to K^m$ is uniquely determined by its restriction $D^n\to K^m\sset F^m$. There are only $2^\kappa$ such functions, hence we can diagonalize against them.

So, let $\{f_\alpha,g_\alpha\}_{\alpha<2^\kappa}$ be an enumeration of all pairs of continuous functions $f_\alpha\colon D^{n_\alpha}\to F^{m_\alpha}$, $g\colon D^{m_\alpha}\to F^{n_\alpha}$ where $n_\alpha>m_\alpha$. For any $\alpha$ and $a\in F^{n_\alpha}$, we put

$$\ob f_\alpha(a)=\lim_{\substack{x\in D^{n_\alpha}\\x\to a}}f_\alpha(x)$$

if it exists, and similarly for $\ob g_\alpha$. For $a\in F^{n_\alpha},b\in F^{m_\alpha}$, we define

$$h_\alpha(a)=b\iff \ob f_\alpha(a)=b\text{ and }\ob g_\alpha(b)=a.$$

Note that $h_\alpha$ is a partial injective function. Then, if $D\sset K\sset F$ and $h\colon K^n\to K^m$ is a homeomorphism, there exists an $\alpha<2^\kappa$ such that $h\sset h_\alpha$.

We construct a strictly increasing sequence $\{K_\alpha:\alpha\le2^\kappa\}$ of subfields of $F$, and an increasing sequence $\{A_\alpha:\alpha\le2^\kappa\}$ of subsets of $F$, such that:

  • $K_\alpha$ and $A_\alpha$ are disjoint, and of cardinality at most $\kappa+|\alpha|$.

  • There is no field $K_{\alpha+1}\sset K\sset F$ disjoint from $A_{\alpha+1}$ such that $h_\alpha$ restricts to a homeomorphism $K^{n_\alpha}\to K^{m_\alpha}$.

Then $K=K_{2^\kappa}$ satisfies the required conditions, hence what remains is to carry out the construction. We put $K_0=D$, $A_0=\varnothing$, and for limit $\gamma\le2^\kappa$, we define

$$K_\gamma=\bigcup_{\alpha<\gamma}K_\alpha,\qquad A_\gamma=\bigcup_{\alpha<\gamma}A_\alpha$$

as usual. For the successor step, assume $K_\alpha$ and $A_\alpha$ have been already constructed. We drop the subscripts from $f_\alpha,g_\alpha,h_\alpha,n_\alpha,m_\alpha$ to simplify the notation, and we say that $K$ is a good field if it is a subfield $K_\alpha\sset K\sset F$ disjoint from $A_\alpha$.

Case 1f: There is a good field $K$ such that $K^n\nsset\dom(\ob f)$. We pick $(a_1,\dots,a_n)\in K\smallsetminus\dom(\ob f)$, and define $K_{\alpha+1}=K_\alpha(a_1,\dots,a_n)$, $A_{\alpha+1}=A_\alpha$.

Case 2f: There is a good field $K$ such that $\ob f[K^n]\nsset K^m$. We pick $(a_1,\dots,a_n)\in K^n$ and $i$ such that $b_i\notin K$ for $(b_1,\dots,b_m)=\ob f(a_1,\dots,a_n)$. We put $K_{\alpha+1}=K_\alpha(a_1,\dots,a_n)$, $A_{\alpha+1}=A_\alpha\cup\{b_i\}$.

Cases 1g, 2g: There is a good field such that $K^m\nsset\dom(\ob g)$, or $\ob g[K^m]\nsset K^n$: similar.

Case 3: There is a good field such that $\ob f\restriction K^n$ and $\ob g\restriction K^m$ are not mutually inverse. Say, $a\ne a'\in K^n$, $b\in K^m$, $\ob f(a)=b$, and $\ob g(b)=a'$. We put the coordinates of $a,a',b$ in $K_{\alpha+1}$.

Case 4: None of the previous cases applies, i.e., $h$ restricts to a bijection $K^n\to K^m$ for every good field $K$. I claim that this is in fact impossible. Recall that $n>m$. Since $|K_\alpha(A_\alpha)|<2^\kappa$, we can find $a_1,\dots,a_n\in F$ algebraically independent over $K_\alpha(A_\alpha)$. Then

$$K_\alpha(a_1,\dots,a_n)\cap A_\alpha=K_\alpha\cap A_\alpha=\varnothing,$$

hence $K=K_\alpha(a_1,\dots,a_n)$ is a good field, thus by assumption, $h(a_1,\dots,a_n)=(b_1,\dots,b_m)\in K^m$. Then $K'=K_\alpha(b_1,\dots,b_m)$ has transcendence degree at most $m$ over $K_\alpha$, hence it is a proper subfield of $K$. However, it is also a good field, hence using the assumption again, $h^{-1}(b_1,\dots,b_m)\in(K')^n$, which implies $K\sset K'$, a contradiction.

A minor final complication is that the previous steps do not ensure that $K_\alpha\subsetneq K_{\alpha+1}$. However, this is easy to fix: by the argument in Case 4, there always exists $a\in F$ transcendental over $K_\alpha$ such that $K_\alpha(a)$ is a good field, hence we can throw it in.

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  • $\begingroup$ Addendum: it follows from mathoverflow.net/a/188628 that an OF of size $2^\kappa$ with a dense subfield of size $\kappa$ exists whenever $\kappa=2^{<\kappa}$. This holds e.g. if $\kappa=\beth_\alpha$ with $\alpha$ limit, hence there are arbitrarily large such fields provably in ZFC. $\endgroup$ – Emil Jeřábek supports Monica Dec 3 '14 at 18:52

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