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Unit sphere arcwise connected?

  1. Dec 29, 2006 #1
    1. The problem statement, all variables and given/known data
    Show that the unit sphere {(x, y, z) : x^2 + y^2 + z^2 = 1} in R^3 is arcwise connected.

    2. Relevant equations

    3. The attempt at a solution
    find a continous map f(t ) such that f( 0) = a, f(1 ) = b. a, b, in R^3 and are on the unit sphere. then show for every t in [0,1] f(t ) is on the sphere. just having trouble finding the right f . Tried f = ta + (1-t)b but only noticed that it is a line segment , so it should be wrong..
  2. jcsd
  3. Dec 29, 2006 #2
    a sphere is convex? I know balls are convex.. just by intuition, a line from a -> b (a, b, on the sphere, the 'shell') is not on the shell of the sphere. .? since {(x, y, z) : x^2 + y^2 + z^2 = 1} , equality is required not <= (i may be wrong)
  4. Dec 29, 2006 #3


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    Just pick any path between the points that doesn't go through zero (eg, the straight line path unless they are diamterically opposed) and project it on the sphere by dividing each vector r(t) by its length. There is a slight technical difference between paths and arcs: arcs can't intersect themselves. But it shouldn't be hard to show this isn't a problem for the projections here.
  5. Dec 29, 2006 #4
    would all the points on the vector ab (a,b on the sphere) be on the sphere?.. i cannot see that and I don't think so..
    I am just following the def of arc-connectness:
    A set S in Rn is arc-connected if any 2 points in S can be joined by a continous curve in S, that is, if for any a, b, in S, there is a continous map f : [0,1] ->Rn s.t. f( 0) = a, f( 1) = b and f(t ) is in S for all t in [0, 1]
  6. Dec 30, 2006 #5

    matt grime

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    No, they're not on the sphere. But StatusX told you how to get round that.
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