Given infinitely many linear inequalities, can we find an algebraic equation for the boundary? Say we have infinitely many linear inequalities cutting out a set of points inside some $\mathbb{R}^n$. Is there any systematic way of translating these into an algebraic equation for the boundary of this set?
For example, fix five points in the plane $x_1,x_2,x_3,x_4,x_5 \in \mathbb{R}^2$, where $x_5$ is inside the quadrilateral defined by the other four points. In other words, there exists infinitely many convex combinations $x_5 = \sum_{i=1}^{4} \mu_i x_i$ writing $x_5$ as a convex combination of the other points (the $\mu_i$ are non-negative and sum to 1). The set of vectors $y \in \mathbb{R}^n$ such that $0 \leq y^T \begin{bmatrix} -\mu_1\\ -\mu_2\\ -\mu_3 \\ -\mu_4 \\ 1\end{bmatrix}$ defines some set of vectors $y \in \mathbb{R}^n$. Can we find an equation for the boundary of this set? Can we even determine if it is algebraic?
 A: For the general case, it depends a lot on the specific inequalities involved. Consider the following two families of inequalities:


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*Let $f_{x_0}(x,y)=(y-x_0^2)-2x_0(x-x_0)$ be the equation of the tangent line to $y=x^2$ at the point $(x_0,x_0^2)$. Then the list of inequalities $f_{x_0}(x,y) > 0$ for all $x_0\in \Bbb R$ describes the region $y>x^2$ which has algebraic boundary.

*Let $f_n(x,y)=(y-n^2)-(2n+1)(x-n)$ be the function which when set equal to zero gives a line connecting the points $(n,n^2)$ and $(n+1,(n+1)^2)$. Then the list of inequalities $f_n > 0$ for all $n\in\Bbb Z$ describes a region which has as it's boundary the line segments connecting $(n,n^2)$ to $(n+1,(n+1)^2)$ for all $n$. This can't be algebraic nor even semialgebraic since it has an infinite intersection with the curve $y=x^2$ without being equal to that curve.
Let's look at a specific example to see if there's any hope. Set $x_1=(1,0)$, $x_2=(-1,0)$, $x_3=(0,1)$, $x_4=(0,-1)$, and $x_5=(0,0)$. The convex combinations of these first four which sum to the last one are given by $\mu_1=\mu_2$ and $\mu_3=\mu_4$, so our inequalities are given by $-\mu_1(y_1+y_2)-\mu_3(y_3+y_4)+y_5 \geq 0$. Up to a linear change of variables, this is just the equation $y_5 \geq \mu_1y_1+\mu_3y_3$.
For any fixed value of $y_5$, the area enclosed by these inequalities is the quadrant in 4-space defined by $y_1\leq y_5/2$ and $y_3\leq y_5/2$. So the solution to all of these simultaneous inequalities is again a quadrant in 5-space, which has a boundary which isn't algebraic (though it is semialgebraic). So the answer to whether the boundary is an algebraic variety is no in general (though the semialgebraic question isn't resolved [but you didn't ask that explicitly either]).
It seems like in general one should concentrate on the extreme values of the various $\mu$ and attempt to reduce from there, but I'm afraid I don't see a general strategy yet. Maybe something along the lines of using convex combinations of the inequalities over some specially chosen (minimal?) selection of $\mu$ would do it, but I'm not sure. If I think of anything else, I'll come back and add to the answer.
