A question about large real closed fields

Question

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 non-empty 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.

Answer 1

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.

Answer 2

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 Löwenheim-Skolem 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 $k^+\le\left|F\right|\le \left(k^+\right)^\lambda=k^+$. It remains to be proven that every subset of $F$ of cardinality at most $\lambda$ is bounded. Choose a well-order $\left(a_\alpha\right)\_{\alpha\in\lambda}$ 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.