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.



A question about large real closed fields
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. 


