If a variety over a real closed field has finitely many points they are singular Let $F$ be a real closed field. Let $X$ be a positive-dimensional algebraic variety over $F$.
If $X$ finitely many $F$-points are they all singular?
 A: Sure. Suppose for contradiction that $P$ is a smooth point. Let $n$ be the dimension of $X$. Consider a projection $X \to \mathbb A^n$ that induces an isomorphism from the tangent space of $X$ at $P$ to the tangent space of $\mathbb A^n$.
This map is etale in a neighborhood of $P$, so it is locally standard etale, i.e. $X$ is locally of the form $\operatorname{Spec} F [x_1,\dots, x_n,y, g^{-1}]/f$ where $f$ and $g$ are polynomials in $x_1,\dots, x_n$, with $f$ monic in $y$ and $\frac{df}{dy}$ a unit.
Let $\overline{x}_1,\dots, \overline{x}_n, \overline{y}$ be the coordinates of $P$. Then since $\frac{df}{dy}(P) \neq 0$, and wlog it is possitive, for $\delta>0$ small we have $f(\overline{x}_1,\dots, \overline{x}_n, \overline{y}+\delta)>0$ and $f(\overline{x}_1,\dots, \overline{x}_n, \overline{y}-\delta)<0$. Then for $\epsilon$ small we have $f(\overline{x}_1+\epsilon,\dots, \overline{x}_n, \overline{y}+\delta)>0$ and $f(\overline{x}_1+\epsilon,\dots, \overline{x}_n, \overline{y}-\delta)<0$ so by the intermediate value theorem there is $y'$ satisfying $y-\delta < y' < y+\delta$ with $f(\overline{x}_1+\epsilon,\dots, \overline{x}_n,y')=0$.
Since this works for infinitely many $\epsilon$, there are infinitely many points, contradicting our assumption.
(I used the definition of real closed field as fields satisfying the intermediate value theorem for polynomials, which is, for example, the 8th one on Wikipedia.)
