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Re: Determining continuity of regions/curves from inequalities

  • To: mathgroup at
  • Subject: [mg67271] Re: [mg67216] Determining continuity of regions/curves from inequalities
  • From: "Carl K. Woll" <carlw at>
  • Date: Thu, 15 Jun 2006 03:25:54 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

Bonny Banerjee wrote:
> Is there an easy way in Mathematica to determine whether the region or curve 
> formed by a system of inequalities is continuous or not?
> For example, the output of some function (e.g. Reduce) might be as follows:
> x>2 && y>0
> which forms a continuous region. Again, the following output
> (x<2 && y<0) || (x>2 && y>0)
> is not continuous. Similarly, for curves.
> Given such a system of inequalities, how to determine whether the 
> region/curve it forms is continuous or not? Or in other words, if I pick any 
> two random points, say P1 and P2, lying on the output curve/region, does 
> there exist a continuous path lying entirely within the output curve/region 
> from P1 to P2?
> Any help will be appreciated.
> Thanks,
> Bonny. 

The function SemialgebraicComponents in the package 
Algebra`AlgebraicInequalities may help. From the help browser:

"The package provides a function for solving systems of strong 
polynomial inequalities in one or more unknowns. To be precise, 
SemialgebraicComponents[ineqs, vars] gives a finite set of solutions of 
the system of inequalities. That is, within the set of solutions, any 
solution can be connected by a continuous path to a solution in the 
finite set. The variable ineqs is a list of strong inequalities, where 
both sides of each inequality are polynomials in variables vars with 
rational coefficients. In other words, SemialgebraicComponents[ineqs, 
vars] gives at least one point in each connected component of the open 
semialgebraic set defined by inequalities ineqs. "


First example: x>2 && y>0

SemialgebraicComponents accepts a list of inequalities:



Only one point is returned, so the region is connected.

Second example: (x<2 && y<0) || (x>2 && y>0)

This example is not a list of inequalities due to the Or. However, in 
this case it is easy to construct an equivalent inequality that 
encompasses the same region:



There are two components, so the region is not connected.

Carl Woll
Wolfram Research

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