The same sort of argument can be used if you're avoiding any countable collection of algebraic relations.

You can use Zorn's lemma and measure theory. Suppose $S$ is a maximal subset of $\mathbb{R}^n$ with no algebraic volumes - one exists by Zorn's lemma. Let $D$ be the set of all points $z$ of $\mathbb{R}^n$ such that $z$ forms a simplex of algebraic volume together with some $n$ points of $S$. Since the set of $n$-tuples of points of $S$ is countable, and the set of algebraic numbers is countable, you see that $D$ is a union of a countable collection of hyperplanes, thus has measure zero. This means that there is a point in the complement to $D\cup S$, so $S$ is not maximal - contradiction!

Whether or not you assume commutativity, you won't in general get connectivity. Take for example maps from the n'th neighborhood of $0$ in $\mathbb{A}^1$ into the n'th neighborhood of $0$ in $Spec(k[x,y]/(xy)$. These mapping spaces are in general at least as complicated as mapping spaces of fixed degree between projective varieties: if $A$ is functions on a neighborhood of the affine cone of a projective variety $X$ and $B$ is functions in a neighborhood of 0 in a veronese twisted affine cone of $Y$ then $Map(B,A)$ goes to $Map(X,Y)$ surjectively by taking blowups of tangent cones.

Ah sorry, thought you were twisting by a character of the Levi (which would not affect the dimension... your current question makes a lot more sense).

Hi David, I mean the category of perfect modules over $T(k)$, which I think it's enough to consider as a DG algebra over $k$. (Correct me if this doesn't make sense).