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](http://at.yorku.ca/p/a/c/a/25.pdf) and a discussion [here](https://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals).)


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.* 



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> is it the **only** ordered field which is not homeomorphic to it's power? 
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Putting this in another way, I want know if this property is a completeness axiom (in the sense discussed for example [here](https://mathoverflow.net/questions/65874/converse-to-banach-s-fixed-point-theorem-for-ordered-fields) or [here](http://arxiv.org/abs/1204.4483)).



Which ordered fields are homeomorphic to their power?
$\let\ob\overline\let\sset\subseteq\let\nsset\nsubseteq\DeclareMathOperator\dom{dom}$ There are many such fields different from $\mathbb R$.



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> **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.
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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](https://mathoverflow.net/a/140962), 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.


