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The a set is open iff its complement is closed?

  1. Mar 15, 2014 #1


    Around the 4 minute mark the lecturer makes this statement, but I am not convinced this is true. I accept that

    (1) if a set is closed, its complement is open.

    but consider the converse.

    Consider an open ball [itex]S[/itex] of some arbitrary radius centered at the origin (in whatever dimension [itex]d[/itex] you want). [itex]S[/itex] is trivially subset of itself, and its complement (in [itex]S[/itex] that is) is [itex]S^c = S \setminus S[/itex] is the empty set [itex]ø[/itex]. But the empty set is open, implying [itex]S[/itex] is closed by (1), and this is a contradiction since we started with choosing [itex]S[/itex] as open.

    What am I missing here? Does it have to do with choosing the complement in [itex]S[/itex] rather than the complement in [itex]\mathbb{R}^d[/itex]?
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Mar 15, 2014 #2

    pwsnafu

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    If you take ##S## as your entire space (which is what you have done), then ##S## is by definition both open and closed in itself. ##S## is not closed relative to the entire ##\mathbb{R}^d##.
     
  4. Mar 15, 2014 #3

    micromass

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    Yes. Here, you regard ##S## as a space in itself. You didn't regard ##S## as the subset of some bigger space.

    In this case, ##S## is indeed open and closed in ##S## (we call that clopen). However, ##S## is open in ##\mathbb{R}^d##, but not closed in ##\mathbb{R}^d##.

    So the difference is the big space you're working in:
    If you're working in ##S##, Then closed in ##S## means that the complement relative to ##S## is open. So ##S\setminus S## is open, which is true.
    If you're working in ##\mathbb{R}^d##, then closed in ##S## means that the complement relative to ##\mathbb{R}^d## is open. So ##\mathbb{R}^d\setminus S## is open, which is true.

    So open and closed are relative notions, depending on the bigger space.
     
  5. Mar 15, 2014 #4
    I see. Thanks for the speedy replies... The names "closed" and "open" are really unfortunate it seems.
     
  6. Mar 15, 2014 #5

    micromass

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    https://www.youtube.com/watch?v=SyD4p8_y8Kw
     
  7. Mar 16, 2014 #6
    I've been away from all this for a long time, but seem to recall that the empty set is vacuously both open and closed.
     
    Last edited by a moderator: Sep 25, 2014
  8. Apr 9, 2014 #7
    Yes, the empty set and the whole space are clopen.
     
  9. Apr 9, 2014 #8

    HallsofIvy

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    In fact, if a set is not "connected" then all "connectedness components" (connected set which are not properly contained in any connected sets) are both open and closed.
     
  10. Apr 9, 2014 #9

    micromass

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    Closed yes. Open no, consider ##\mathbb{Q}##. To have the connected components open you need the notion of locally connectedness: http://en.wikipedia.org/wiki/Locally_connected_space
     
  11. Apr 9, 2014 #10

    D H

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    Godwin's law meets topology!
     
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