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I am given the following semi-algebraic set: $$ \{ x= \begin{bmatrix} x_1,x_2,x_3,x_4,x_5 \end{bmatrix} ^T\in\mathbb{R}^5| -2 \leq (1 - x_2x_3)x_5 - 2x_4x_3x_1 \leq 1.5,\\ ~ -1 \leq x_1 \leq 30,\\ 0 \leq x_2 \leq 3,\\ ~-1/50 \leq x_3 \leq 1/50,\\ ~ -13 \leq x_4 \leq 13,\\ ~ -22 \leq x_5 \leq 22 \}, $$ of which I would like to know all full-dimensional, connected components. For this, I use Mathematica:

SetSystemOptions["InequalitySolvingOptions" -> "CADCombineCells" -> Automatic];
eqns = -2 <= (1 - x2*x3)*x5 - 2*x4*x3*x1 <= 15/10 && -1 <= x1 <= 30 && 
0 <= x2 <= 3 && -1/50 <= x3 <= 1/50 && -13 <= x4 <= 13 && -22 <= x5 <= 22;

{res, trash} = GenericCylindricalDecomposition[eqns, {x1, x2, x3, x4, x5}];
res

As a result, I get:

-1<x1<30&&0<x2<3&&-(1/50)<x3<1/50&&-13<x4<13&&(2-2 x1 x3 x4)/(-1+x2 x3)<x5<(-3-4 x1 x3 x4)/(-2+2 x2 x3)

I get the same result using:

 CylindricalDecomposition[eqns, {x1, x2, x3, x4, x5}]

instead of the generic decomposition. Now my questions:

  1. Can I conclude from these findings that the set consists of only a single connected component?
  2. Can I conclude that there are no lower-dimensional components?
  3. Does Mathematica's "connected" mean "simply connected", i.e., there are no holes in the component?

Please note that, as an engineer, I have only a rough idea of the underlying algebraic geometry.

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  • $\begingroup$ This is rather a math question than a Mathematica question. Ask it at math.stackexchange.com . $\endgroup$ – user64494 Oct 8 '18 at 12:54

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