Let $\mathbb{A}\_{\mathbb{R}}^n$ be $\mathbb{R}^n$ endowed with the Zariski topology, where closed sets are algebraic sets (in $\mathbb{R}^n$) defined by real polynomials.


Suppose $V \subseteq \mathbb{A}\_{\mathbb{R}}^n$ is an irreducible affine variety. Let $U$ be an open (with respect to the usual topology) ball $U$
around a non-singular point of $V$ and of small enough radius. 


Does it then follow that the Zariski closure of $(V \cap U)$ is $V$?
I thought it should be true (maybe not?), but I was wondering how I can show this.
Any comments are appreciated. Thank you! 



Question about taking the Zariski closure in $\mathbb{A}_{\mathbb{R}}^n$
As noted in [this answer](https://mathoverflow.net/a/297013/82179) to a previous question of yours, for any subvariety $W \subseteq \mathbb A\_\mathbb R^n$, we have
$$\dim\_{\mathbb R} W(\mathbb R) \leq \dim W$$
(where $\dim W$ denotes the dimension in the sense of scheme theory, written there as $\dim\_\mathbb C W(\mathbb C)$), with equality if $W$ has a smooth $\mathbb R$-point. In particular, if $W \subseteq V$ is the Zariski closure of $V \cap U$, then applying the above to both $V$ and $W$ gives
$$\dim\_\mathbb R W(\mathbb R) \leq \dim W \leq \dim V = \dim\_\mathbb R V(\mathbb R).$$
But $W(\mathbb R)$ contains the full-dimensional subset $V(\mathbb R) \cap U$ of $V(\mathbb R)$, hence all dimensions must be equal. Since $V$ is irreducible and $\dim W = \dim V$, this forces $W = V$. $\square$


