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Hausdorff Space

  1. May 26, 2010 #1
    We have a equivalence relation on [0,1] × [0,1] by letting (x_0, y_0) ~ (x_1, y_1) if and only if x_0 = x_1 > 0... then how do we show that X\ ~is not a Hausdorff space ?
  2. jcsd
  3. May 27, 2010 #2
    Wait, so (0,0) is not equivalent to itself? Then it's not an equivalence relation?
  4. May 27, 2010 #3
    I think topologists usually don't bother explicitly describing eqivalence classes with a single member, so in this case each {(0,y)} would be a singleton equivalence class. (What can you say about open sets around {(0,y)} and {(0,y')} in the quotient topology, where y and y' are distinct?)
    Last edited: May 27, 2010
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