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RP^n as a topological space

  1. Jul 16, 2009 #1
    so first let's take RP^2. I have a little trouble grasping why we can put a subspace topology on it. So RP^2 is the set of all lines through the origin in R^3. So if we take some subset S of RP^2 and the if set of points in R^3 which is the union of these lines in S is open then we can say we take the intersection of S and the open subset of R^3 and get a subspace topology. However, apparently you can only get the indiscrete topology? the reason i read for this was because the union would contain 0 but you can't have an open ball around 0. Let's say you take a single line as your subset S. you can't put an open ball around any of the points right? you won't necessarily have some delta in the x, y, z directions? how come the argument says you can't put an open ball around 0, but you can put it around the other points?
    @_@
     
  2. jcsd
  3. Jul 19, 2009 #2
    The basic reason is that any neighborhood of zero contains a point on every line through the origin.
     
  4. Jul 20, 2009 #3
    so if we take some point P which is not 0, what would P's neighborhood consist of? and what would happen to the neighborhood when we take subsets of RP^n
     
  5. Aug 10, 2009 #4
    In the definition of projective space the lines through the origin are topologized by the set that they determine when intersecting the sphere. A set of lines is open if its intersection with the sphere is open in the sphere.

    A point in projective space is a line. A point on this line is not a point of projective space. Zero is not a point nor is any other point on the line.
     
  6. Aug 10, 2009 #5

    Office_Shredder

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    The "correct" way to modify the quotient topology to work on RPn is to look at the unit sphere of Rn+1 and take the open sets on that, and match them with the sets of lines passing through the points of the open sets in Rn+1. This is the topology you're looking for in RPn
     
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