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Locally path connected implies path connected

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    We say the metric space X is locally path connected (lpc) if all balls are path connected sets.

    Suppose X is lpc and that E is an open and connected subset of X. Prove that E is path connected.

    2. Relevant equations
    A set is open if every point is an interior point. A set is connected if it cannot be written as the union of two open, disjoint, non-empty sets. A set is path connected if given any two points x,y in the set, there exists a path between them.


    3. The attempt at a solution
    None really. My professor suggested that for a fixed x, I consider the set Y of all y that are path connected to x and show that this set is all of E by somehow showing that if it were not, I could disconnect the space using Y and Y complement.

    I just don't see the direction that this problem could take. Is it long? Is it short but requires ingenuity? What should I try to do? Where am I trying to end up?

    Help, it's hard!
     
  2. jcsd
  3. Nov 8, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    It's not hard and doesn't require any stunning ingenuity. You just have to think clearly. Can you show Y is open? The main ingredient you need is that if a,b and c are three points, and a is path connected to b and b is path connected to c then a is path connected to c. Now can you show E-Y (the set of all points in E that aren't in Y) is also open? That's your disconnection.
     
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