Countable union of non Zariski-dense homomorphisms Let $F_k$ be a free group in $k>1$ letters, and $G$ a semi-simple algebraic group defined over reals $\mathbb{R}$. Consider the representation variety Hom$(F_k,G(\mathbb{R}))$. The points of this variety are the homomorphisms $\phi: F_k \to G(\mathbb{R})$. Consider the set of homomorphisms with non Zariski-dense image, i.e. those homomorphisms $\overline{\phi(F_k)}$ is a proper subgroup of $G(\mathbb{R})$, where the closure is considered in Zariski topology. Denote the subset of such homomorphisms by $\mathcal{F}$.
My question is whether $\mathcal{F}$ is Zariksi-dense in Hom$(F_k,G(\mathbb{R}))$ ?
 A: Yes: actually $\mathcal{F}$ is Zariski-closed in $G^k$. (And since $\mathcal{F}\neq G^k$ as soon as $k\ge 1$ and $G\neq\{1\}$, we deduce in this case that $\mathcal{F}$ is not Zariski-dense.)

All this can be performed over an algebraic closure, so in the following I never suppose anything to be defined over the reals.
Let $(V,\pi)$ be an irreducible representation of $G$ (of dimension $d_V$). Let $U_V$ be the set of $(g_1,\dots,g_k)\in G^k$ acting irreducibly on $V$. Then it means that the subalgebra generated by $\pi(g_1),\dots,\pi(g_k)$ contains $d_V^2$ linearly independent elements. This is a Zariski open condition.
By Chevalley's theorem, every proper subgroup is a point stabilizer in some representation. We perform this with every maximal Zariski-closed subgroup: these are finitely up to conjugacy (indeed they are either parabolic, or reductive; in the second case this means the normalizer of a semisimple subgroup, and there are finitely many semisimple subgroups up to conjugacy.) We thus get representations $V_1,\dots,V_n$ corresponding to maximal subgroups $M_1,\dots,M_n$. A subgroup contained in a conjugate of $M_i$ is non-irreducible on $V_i$. Hence if $(g_1,\dots,g_k)$ is in $U=\bigcap U_{V_i}$, then it generates a Zariski-dense subgroup and conversely being outside $U_{V_i}$ implies failure of Zariski-density. Thus $\mathcal{F}$ is the complement of $U$, and thus is Zariski-closed.
For $k\ge 2$ and $G\neq 1$ there are indeed Zariski-dense 2-generated subgroups and in this case $\mathcal{F}\neq G^k$, so $\mathcal{F}$ is not Zariski-dense. 
