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Help with proof

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data
    I need help proving that given two surfaces
    any two points of surface X can be joined by a curve in X, while the same is not true of Y.

    2. Relevant equations
    X={(x,y,z):x^2+y^2-z^2=1} and Y={(x,y,z):x^2+y^2-z^2=-1}

    3. The attempt at a solution
    I'm not sure where to start. If the problem had said line instead of curve, I think I could do a proof by contradiction, but as it is I don't know.
  2. jcsd
  3. Jan 18, 2009 #2


    Staff: Mentor

    A start would be to sketch graphs of the two surfaces. I suspect that once you know what they look like, it will be easier to prove what you need to prove.

    If the problem had said line instead of curve, it would have been even harder to prove, since the line would also have to lie on the surface, not just go from one point to another.
  4. Jan 18, 2009 #3


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    Homework Helper

    You might notice that the equation for the surface Y has no solutions for z=0. But it does for z>1 or z<-1. Is that enough of a hint to get you started?
  5. Jan 19, 2009 #4
    I know that there's a gap between z=-1 and z=1, but I'm not sure how to prove the general case that not all points can be connected by a curve in Y. So, I can see the solution, but I don't know how to argue it.
  6. Jan 19, 2009 #5


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    Staff Emeritus
    Science Advisor

    As Mark44 said, graphing them would help. For surface X, [itex]z^2= x^2+ y^2+ 1[/itex] and in cylindrical coordinates that is [itex]z^2= r^2- 1[/itex] or [itex]r^2- z^2= 1[/itex]. In the "r,z" plane, that is a hyperbola with r axis as axis of symmetry. Rotating around the z-axis, then, we have a "hyperboloid of one sheet". For the surface Y, [itex]z^2= x^2+ y^2+ 1[/itex] or [itex]z^2= r^2+ 1[/itex] which, in the 'r,z-plane" is z^2- r^2= 1, a hyperbola with the z axis as axis of symmetry. Rotating around the z-axis, we have a "hyperboloid of two sheets".

    Given two points on surface X, say [itex](x_0,y_0, z_0)[/itex] and [itex](x_1,y_1,z_1)[/itex] we can draw the straight line from [itex](x_0,y_0,z_0)[/itex] to [itex](x_0,y_0,z_1)[/itex], then a straight line from that point radially inward to [itex](x_3,y_3,z_1)[/itex] where [itex]y_3/x_3= y_0/x_0[/itex] and [itex]\sqrt{x_3^2+ y_3^2}= \sqrt{x_1^2+ y_1^2}[/itex]. Finally, a circle with center at [itex](0,0,z_1)[/itex] and radius [itex]\sqrt{x_1^2+ y_1^2}[/itex] will take us to [itex](x_1,y_1,z_1)[/itex] while staying on the surface.

    For surface Y, [itex]z^2= x^2+ y^2+ 1[/itex], or [itex]z= \pm\sqrt{x^2+ y^2+ 1}[/itex] which makes it clear that there is no point on the surface with [itex]-1< z< 1[/itex]. Any point [itex](x_0,y_0,z_0)[/itex] on the surface, with z> 0, can be written as [itex]z_0= \sqrt{x_0^2+ y_0^2+ 1}[/itex]. Another point on the surface is [itex](x_0, y_0, -z_0). Since no point on the surface has z between -1 and 1, it is impossible to connect those two points by a continuous curve.
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