Re: Determining continuity of regions/curves from inequalities

*To*: mathgroup at smc.vnet.net*Subject*: [mg67241] Re: [mg67216] Determining continuity of regions/curves from inequalities*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 14 Jun 2006 06:28:51 -0400 (EDT)*References*: <200606130506.BAA23751@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 13 Jun 2006, at 14:06, 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. > In general the answer is complicated, but in cases like the above you can tell how many components there are buy looking at how many Or's appear in the CylindricalDecomposition of your set of inequalities. For example consider ss = CylindricalDecomposition[x^3 + y^2 + z^4 < x^5, {x, y, z}] (-1 < x < 0 && -Sqrt[x^5 - x^3] < y < Sqrt[x^5 - x^3] && -(x^5 - x^3 - y^2)^(1/4) < z < (x^5 - x^3 - y^2)^ (1/4)) || (x > 1 && -Sqrt[x^5 - x^3] < y < Sqrt[x^5 - x^3] && -(x^5 - x^3 - y^2)^(1/4) < z < (x^5 - x^3 - y^2)^(1/4)) so there are two components. Take the point (2,0,0}. We have ss[[1]]/.Thread[{x,y,z}->{2,0,0}] Out[11]= False In[12]:= ss[[2]]/.Thread[{x,y,z}->{2,0,0}] Out[12]= True So (2,0,0) lies in the first component. Now consider (-1/2,0,0) ss[[1]] /. Thread[{x, y, z} -> {-2^(-1), 0, 0}] True ss[[2]] /. Thread[{x, y, z} -> {-2^(-1), 0, 0}] False so (-1/2,0,1} lies in the second component. The two cannot be connected by a curve lying within the region satisfying the inequality. You can see the two components as follows: <<Graphics`InequalityGraphics` InequalityPlot3D[x^3+y^2+z^4<x^5,{x,-3,3},{y},{z}] Andrzej Kozlowski

**References**:**Determining continuity of regions/curves from inequalities***From:*"Bonny Banerjee" <banerjee@cse.ohio-state.edu>