Let $RCF=Th(\mathbb R ,+,\cdot ,<,0,1)$. We know RCF is $o$-minimal, i.e every definable set is a finite union of points and intervals with endpoints.
I want to show that definable set in that theory is either finite or in the cardinality of the model or at list not countable.And with that i will be able to conclude that RCF has no Vaughtian pair.
I had an attempt but i think it wont work... What i tried is to say that in a model $M$ of RCF i can take a given interval, say $(a,b)$ and then, i can elurge it in a way that i keep the same cardinality and include any arbitrary member i want, by finding some member $c$ and taking $(a-c,b-c)$. and if i assume towards contradiction that RCF has a Vaughtian pair, then in prticular, there is a model of RCF of size $\aleph_1$ with a countable definable set, and by $o$-minimality we have a countable interval $(a,b)$, But then also $(a-c,b+c)$ is countable for all $c$, and then i tried to say that in some point we get a interval of cardinality of $\aleph_1$ from regularity or some thing. But was unable.I don't think it will work
Edit: my friend had an idea that one can show that if two models of RCF has a common interval, then they are equal. To do that what i tried was to assume $M,N\vDash RCF$ and $I^M =I^N$ if say, there was $c\in |N|\setminus |M|$, and $I=(a,b)$, then we can show the existence of some $d\in |M|$ s.t $N\vDash d>c$ and $d$ is "close enough to $c$ so that $c\in (a+d,b+d)$. i am still not sure how to solve the "close enough" problem...
Let $R$ be a real closed field. Let's say an interval is bounded if neither of its endpoints are $\pm\infty$. Any two bounded intervals $(a,b)$ and $(c,d)$ have the same cardinality. Indeed, $t\mapsto \frac{d-c}{b-a}(t-a) + c$ defines a bijection between them.
Now the map $f(x) = \frac{x}{x^2-1}$ is a bijection from the interval $(-1,1)$ to the interval $(-\infty,\infty)$ in $R$. To see that $f$ is surjective (which is all you need for the cardinality argument), note that for any $b\in R$, the polynomial $p(x) = b(x^2-1) - x$ has $p(-1) = 1$ and $p(1) = -1$, so by the intermediate value theorem for polynomials, $p$ has a zero $a$ in the interval $(-1,1)$, and $f(a) = b$.
This shows that every bounded interval $(a,b)$ has the same cardinality as $R$.