Image of semialgebraic set is a semialgebraic set 
Let $f: \mathbb{R}^n \to \mathbb{R}^m$, where $m\le n$, be a linear map that maps all points with rational coordinates to the points with rational coordinates. Prove that the $f(A)$ is semialgebraic if $A$ is semialgebraic set.

From Tarski–Seidenberg theorem we know that projection is a semialgebraic set, but how to deal with a linear map, which can not be a projection?

Consider the graph $\Gamma_f\subset \Bbb R^n\times \Bbb R^m$. This can be written as the zero set of a collection of linear equations, so it's semialgebraic. Next, intersect $\Gamma_f$ with the set $A\times \Bbb R^m$, which is semialgebraic as the product of semialgebraic sets is semialgebraic. But this intersection is exactly $(a,f(a))$ for $a\in A$, so by projection to $\Bbb R^m$ we see that we get $f(A)$ as a semialgebraic set.
This same procedure works in general to show that the image of a semialgebraic set under a semialgebraic map (defined as a map with semialgebraic graph) is again semialgebraic.
