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Isn't the circle connected?

  1. Jul 30, 2009 #1
    This is from Hatcher's Algebraic Topology, page 30.

    I thought that the circle, S1 is path connected. How then can it be decomposed into the disjoint union of open sets? Furthermore, how can two disjoint open arcs in S1 be the have S1 as their union? What happened to the boundary of the two open sets?

    Thanks.
     
  2. jcsd
  3. Jul 30, 2009 #2
    It is. Therefore, it's connected.

    It can't. Reread the snippet carefully. You take a cover of S1, then run it backwards through p. The result is two disjoint open sets in R (not S1!).

    You might think "well, p is continuous and connectedness is a topological property?" But that doesn't matter because p^-1 is not continuous.
     
  4. Jul 30, 2009 #3

    George Jones

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    Also note that a disjoint union of open sets A and B does not, in general, require that A and B be disjoint, although, as Tac-Tics has noted, in this case the inverse image of an open arc in S1 is the union of disjoint open subsets of R.
     
  5. Jul 31, 2009 #4
    But p was defined to be p(s)=(cos 2*pi*s, sin 2*pi*s). Doesn't that mean that the preimage of a cover in S1 must cover an interval in R? Also, it says that each p-1(U_alpha) was mapped homeomorphically to U_alpha by p. Doesn't that mean that connectedness must be preserved by the inverse function?

    <quote>Also note that a disjoint union of open sets A and B does not, in general, require that A and B be disjoint</quote>

    That makes sense! So that means that the use of disjoint here is redundant since when you take a union, you never choose the same element twice. Am I right?
     
  6. Jul 31, 2009 #5

    George Jones

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    No, it doesn't say this. :smile:
     
  7. Jul 31, 2009 #6
    I'm probably interpreting this wrongly but doesn't it say:

    disjoint union of open sets each of which is mapped homeomorphically
    onto U _alpha by p.

    Thanks for being so patient.
     
  8. Jul 31, 2009 #7

    George Jones

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    "each of which" refers to the individual sets whose union is being taken.

    If

    [tex]p^{-1} \left[ U_\alpha \right] = A_1 \cup A_2 \cup ...,[/tex]

    then each [itex]A_i[/itex] is mapped homeomorphically onto [itex]U_\alpha[/itex].
     
  9. Jul 31, 2009 #8
    You're right! That makes so much more sense now! The disjoint open sets would then just be the periodic repeats of the arcs. I need to read more carefully. Thank you so much - you've made my day! :)
     
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