I want to use the following statement concerning varieties, but I do not know why it is true.
Claim. Let $V \subset \mathbb{C}^n$ be a variety defined over $\mathbb{Q}$. Then the set $V \cap \bar{\mathbb{Q}}$ is dense in $V$ with respect to the Hausdorff topology (not the Zariski topology).
Here, $\bar{\mathbb{Q}}$ denotes the algebraic closure of the rationals $\mathbb{Q}$.
It was pointed out to me that one can show this using the so-called Tarski-Seidenberg Principle, in particular using Proposition 5.3.5 of Real Algebraic Geometry by Bochnak, Coste, Roy.
Let $R$ be a real closed field, $A\subset R^m$ and $B\subset R^n$ semialgebraic sets, and $f:A\to B$ a semialgebraic map with graph $G\subset A\times B$. Let $K$ be a real closed extension of $R$, and denote the extension of a semialgebraic set $S$ defined over $R$ to $K$ as $S_K$.
Proposition 5.3.5
i) The semialgebraic set $A$ is open (resp. closed) in $R^m$ iff $A_K$ is open (resp. closed) in $K^m$. More generally, $clos(A_K)=(clos(A))_K$.
ii) The semialgebraic mapping $f$ is continuous iff $f_k$ is continuous.
I do not see how my statement follows. Does anyone have experience with applying this principle to this kind of situation? Or can you think of another approach leading to a proof of the claim?
A proof originally developed in the comments:
Let $\Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $\Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $\Bbb C^n$ with $\Bbb R^{2n}$, and identifies the $\overline{\Bbb Q}$ points of $V$ with the $\Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $\Bbb C^n$ with $\Bbb R^{2n}$.
From here, if one can show that the $\Bbb R_{alg}$ points of a semialgebraic set $S\subset \Bbb R_{alg}^n$ are dense in the $\Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $\Bbb R_{alg}$. Then the problem reduces to showing that the $\Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $\Bbb R$ points of $(0,1)^d$, which is clear as $\Bbb Q\subset \Bbb R_{alg}$ and the $\Bbb Q$ points are clearly dense in $(0,1)^d$.
This feels a little inelegant, but it works.
Alternatively, picking up from the point where we need to prove that the $\Bbb R_{alg}$ points are dense in the $\Bbb R$ points: Suppose there was a point in $S(\Bbb R)$ which was not in the closure of $S(\Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(\Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $\Bbb R_{alg}$ but semialgebraically disconnected over $\Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)