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The Tarski–Seidenberg theorem asserts that the projection of a semialgebraic set is also a semialgebraic set. My question is whether this is provable in constructive mathematics.

First, let me formalize what this means constructively. Let $\phi(x_1,\dots,x_n,c_1,\dots,c_m)$ be a quantifier-free formula in the language with $+$, $\cdot$, and $<$. Then $$\{(x_1,\dots,x_n) \in \mathbb R^n : \phi(x_1,\dots,x_n,c_1,\dots,c_m)\}$$ is a semialgebraic set for any $c_1,\dots,c_m \in \mathbb R$ ($c_1,\dots,c_m$ are called parameters or constants). This is the same as the usual definitions when using classical mathematics.

For any set $S \subseteq R_{n+1}$, its projection is $$\{(x_1,\dots,x_n) \in \mathbb R^n : \exists x_{n+1} \in \mathbb R. (x_1,\dots,x_n,x_{n+1}) \in S \}$$

The Tarski–Seidenberg theorem states that if S is semialgebraic, its projection is semialgebraic.

We might also consider if S's co-projection $\{(x_1,\dots,x_n) \in \mathbb R^n : \forall x_{n+1} \in \mathbb R. (x_1,\dots,x_n,x_{n+1}) \in S \}$ is semialgebraic. Classically the Tarski–Seidenberg theorem implies this too, but constructively the analogous statement talking about co-projections is different. I'll call this the "co-Tarski–Seidenberg theorem" and am interested in its constructive provability as well.

If they are not constructively provable, I am also curious of how strong they are. For example, does the Tarski–Seidenberg theorem or the co-Tarski–Seidenberg theorem imply the fundamental theorem of algebra.


Classically we can prove that the semialgebraic sets on $\mathbb R^n$ form a Boolean algebra, but constructively we can only show that they form a Heyting algebra.

If we restrict our attention to the algebraic real numbers, I'm pretty sure the Tarski–Seidenberg theorem is constructive since working with the algebraic numbers isn't much different constructively (in particular, they have decidable equality). In fact, I think we can represent Tarski–Seidenberg theorem over the algebraic numbers as a $\Pi^0_2$ statement that can thus be made constructive using the Friedman translation.

However, working with $\mathbb R$ there are significant differences that probably break the original proof. For example the sets $\{x \in \mathbb R: \lnot (x = 0)\}$ and $\{x \in \mathbb R : (x > 0) \lor (x < 0)\}$ are semialgebraic sets that are obviously equal classically but can't be proven equal constructively (but could be proven equal if we replace $\mathbb R$ with the algebraic numbers).

(Also see Is it decidable whether a statement about reals (in the language of ordered rings) is constructively provable? for a closely related question. If both the Tarski–Seidenberg theorem and the co-Tarski–Seidenberg theorem are true, the sentences considered in that question could at least be reduced to the inhabitedness of semialgebraic sets over $\mathbb R^0$.)

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    $\begingroup$ May I suggest that before we even attempt to think about the (likely very delicate) question of constructive quantifier elimination for the reals and/or for real-closed field (and to what extent $\mathbb{R}$ is real-closed or what this even means constructively), the first step should be to fully understand the analogous but simpler questions about $\mathbb{C}$ (and the language consisting of $+,\times,\#$ where $\#$ means “apart”) and Chevalley's theorem about projection of algebraically constructible sets. $\endgroup$
    – Gro-Tsen
    Commented Jun 21 at 18:04
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    $\begingroup$ In fact, if I recall correctly, it is not provable in constructive mathematics without Choice that every complex number (where $\mathbb{C} := \mathbb{R}[i] := \mathbb{R}[X]/(X^2+1)$ where $\mathbb{R}$ denotes the Dedekind reals) has a square root. So the first projection of the set of $(z,w) \in \mathbb{C}^2$ with $z=w^2$ is likely to be problematic. Translated back to the reals, this means that the projection on the first two coordinates of the set of $(x,y,u,v) \in \mathbb{R}^4$ such that $x = u^2-v^2$ and $y=2uv$ might be a counterexample to what you're asking. $\endgroup$
    – Gro-Tsen
    Commented Jun 21 at 18:13
  • $\begingroup$ @Gro-Tsen hmm, here's an even trickier one in the co-projection case: $\forall x. x = y \lor x ≠ y$. No idea how you'd express that as a quantifier-free formula. $\endgroup$ Commented Jun 21 at 20:00
  • $\begingroup$ Maybe it would make sense to think about a slightly more general notion of 'quantifier-free formula' where you allow arbitrary propositional constants. (This idea has shown up in something else I was thinking about.) While the actual truth value of $\forall x. x = y \lor x ≠ y$ is independent, we know it doesn't depend on $y$ constructively. $\endgroup$ Commented Jun 22 at 0:35
  • $\begingroup$ @JamesEHanson I'm also holding out hope that the projection case works if you assume the fundamental theorem of algebra (then $\exists w. z = w^2$ reduces to $\top$ for complex numbers). $\endgroup$ Commented Jun 25 at 18:39

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