A real closed field can be ordered in one and only one way, and is therefore provided with a unique order topology. Given any infinite cardinal number k, does there always exist a real closed field F (whose cardinal number is greater than k), such that no nonempty subset of F having a cardinal number not greater than k has a limit point in F? The criterion for a point of F to be a limit point of a subset of F is dtermined by the order topology of F. Since we are dealing here with arbitrarily large cardinal numbers, let us assume that we are working within the set theory ZFC.

$\begingroup$ If I understand your question correctly, won't the ultrapower of $\kappa^+$many copies of $\mathbb{R}$ over a regular ultrafilter work? $\endgroup$ – Noah Schweber May 18 '13 at 19:03

$\begingroup$ See related question: mathoverflow.net/questions/72612/… $\endgroup$ – Joel David Hamkins May 18 '13 at 19:03

$\begingroup$ Any $\kappa^+$saturated realclosed field has the property you asked for. Since you allow the field to be significantly larger than $\kappa^+$ if necessary, there's no difficulty producing such fields. $\endgroup$ – Andreas Blass May 18 '13 at 19:12
If $\delta$ is the cofinality of an ordered field $F$, that is, the size of the smallest unbounded subset of $F$, then every point of $F$ fills a cut of type $(\delta,\delta)$. In other words, every point in $F$ is the limit of an increasing $\delta$ sequence from below and a decreasing $\delta$ sequence from above. One can see that this is true of $0$ by inverting the elements of a strictly increasing positive unbounded $\delta$ sequence; and then one can translate this sequence from $0$ to any other point for the general conclusion.
It follows that any set with a limit point must have size at least $\delta$, and consequently any set of size less than $\delta$ has no limit points in $F$.
It is easy to make fields of any desired cofinality, just by forming a chain of elementary extensions of that length, adding new points above at each step.

$\begingroup$ The extensions don't have to be elementaryany field extensions will do, as long as they get strictly taller each step. If you extend $\delta$ many times and $\delta$ is regular, then you will get a field of cofinality $\delta. $\endgroup$ – Joel David Hamkins May 18 '13 at 20:33

$\begingroup$ Many thanks for this helpful information. If (x(1),y(1)) and (x(2),y(2)) are elements of F^2, then the formula: ((x(2)x(1))^2)+((y(2)y(1))^2)^(1/2) satisfies all the axioms for a metric on the "space" F^2, although the the resulting "distance" is a nonnegative element of F and not necessarily a nonnegative real number. Of course if F is sufficiently large, then F^2 is not metrizable (in the classical sense) and although it is dense in itself, it is not "complete". $\endgroup$ – Garabed Gulbenkian May 19 '13 at 14:56

$\begingroup$ My question about limit points arises in trying to work out whether and if so, howit might be possible to define some sort of "completion" for F^2 that still allows the "metric" that we have defined to exist. $\endgroup$ – Garabed Gulbenkian May 19 '13 at 15:07

$\begingroup$ In that case, you may be interested in: mathoverflow.net/questions/10870/…. $\endgroup$ – Joel David Hamkins May 19 '13 at 15:59

$\begingroup$ Joel, in reference to that question: Given any metric space Y (in the classical sense) and a set R* of standard and nonstandard real numbers (which I assume is some real closed field like F in my question) how does one define the nonstandard analogue Y* of Y? If d,d* are the respective distance funcions of Y,Y*how does one define d* when d is given? Many standard metrics involve functions which are exponential or inverse trigonometric and which could cause difficulties when one tried to extend them to larger real closed fields. $\endgroup$ – Garabed Gulbenkian May 20 '13 at 18:38
Notice: I thought you would require the field to be of cardinality $k^+$. In that case the proof uses the generalised continuum hypothesis. But without this restriction it is completely unnecessary. The construction is similar to the construction of “the” hyperreals (you might get “them” in the case $\kappa=\omega$).
I can prove it assuming the generalised continuum hypothesis. By the upwards LöwenheimSkolem theorem there is always a real closed field of cardinality $k^+=2^k$. By induction we can now prove that for every cardinal $\lambda\le k$ there exists a real closed field of cardinality $k^+$ such that every subset of cardinality at most $\lambda$ is bounded. Obviously every finite subset of such a field is bounded. Now consider an infinite cardinal $\lambda\le k$ and assume that for every $\mu<\lambda$ there exists a real closed field $F_{\mu}$ of cardinality $k^+$ such that all subsets less than or equal $\mu$ are bounded and $F_\mu\subset F_\nu$ for $\mu<\nu<\lambda$ and $F_\mu=F_\nu$ for $\mu,\nu$ finite. Now choose an ultraproduct $F$ of all the $F_{\left\alpha\right}$ indexed by ordinals $\alpha\in\lambda$ with respect to an ultrafilter which contains all complements of sets of smaller cardinality than $\lambda$. All the $F_\mu$ for $\mu<\lambda$ can be embedded into $F$. The cardinality of $F$ can be approximated by $k^+\le\leftF\right\le \left(k^+\right)^\lambda=k^+$. It remains to be proven that every subset of $F$ of cardinality at most $\lambda$ is bounded. Choose a wellorder $\left(a_\alpha\right)\_{\alpha\in\lambda}$ where each $a_\alpha$ represents an element of $F$ given by a sequence. There exists an upper bound $b$: We just have to choose $b\_\beta$ to be an upper bound of all $(a\_\gamma)\_\beta$ for $\gamma\le\beta$. These upper bounds exist because the set $\left\{(a\_\gamma)\_\beta\mid\gamma\le\beta\right\}$ is contained in $A_{\left\beta\right}$ and has cardinality than or equal to $\left\beta\right$. QED.

$\begingroup$ You don't need the GCH to build fields of your desired cofinality and size. Once you make a taller field at all, you can cut down to a taller one of the same cardinality as the base field by means of the LowenheimSkolem theorem. The union of the chain of extensions will then have the desired cofinality. $\endgroup$ – Joel David Hamkins May 18 '13 at 19:55

$\begingroup$ You are right, I did not notice that it is easy to guarantee the desired confinality using some formulas in predicate calculus. The proof is much too complicated. $\endgroup$ – The User May 18 '13 at 20:02

$\begingroup$ Is it complicated? To make a field $F$ of cofinality $\delta$, a regular cardinal, one should just form $F$ as $\bigcup_{\alpha\lt\delta} F_\alpha$, where $F_{\alpha+1}$ has some points on top of every element of $F_\alpha$. Such a field $F_{\alpha+1}$ exists by the compactness theorem. These new points are now unbounded in $F$, and any set of size less than $\delta$ is contained in some $F_\alpha$ and hence is bounded, since $\delta$ is regular. $\endgroup$ – Joel David Hamkins May 18 '13 at 20:08

$\begingroup$ My proof is much too complicated, not yours, sorry for the confusion $\endgroup$ – The User May 18 '13 at 20:26