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Rudin's Theorem 2.27

  1. Jan 2, 2013 #1
    Prove the closure of E in a Metric Space X is closed. (page 35)

    Rudin states:

    if p∈X and p∉E then p is neither a point of E nor a limit point of E..

    Hence, p has a neighborhood which does not intersect E. (Great)

    The compliment of the closure of E is therefore open. WHY? I don't see it...

    BTW, I know there are different ways to proving this, but I want to understand the last line jump. Thanks.
     
    Last edited: Jan 2, 2013
  2. jcsd
  3. Jan 2, 2013 #2
    What is the definition of open? What has Rudin just shown about an arbitrary point in X?
     
    Last edited: Jan 2, 2013
  4. Jan 2, 2013 #3

    micromass

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    What is your definition of "open"?
     
  5. Jan 2, 2013 #4
    E is Open if every p in E is an interior point (meaning there exists a neighborhood of p that is in E)
    The problem we should say that the complement of E is open, not the complement of the closure of E.
     
  6. Jan 2, 2013 #5
    I guess since the intersection of N(p) and E is empty then no point q of N(p) can be a limit point of E as this would mean every neighborhood of q will contain an infinite number of points in E. Hence the intersection of N(p) and "closure of E" is empty.

    Is this correct?
     
    Last edited: Jan 2, 2013
  7. Jan 2, 2013 #6

    lavinia

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    If any point not in E has an open neighborhood that does not intersect E then by definition the complement of E is open.
     
    Last edited: Jan 3, 2013
  8. Jan 2, 2013 #7
    my bad, I forgot to add the word: "closure" in the last line of the proof. I just re-read it.

    This is what is confusing me:

    FROM: Hence, p has a neighborhood which does not intersect E.

    We get: The compliment of the closure of E is therefore open.
     
  9. Jan 3, 2013 #8
    We have just shown that p is in the complement of the closure of E (call it A). We also showed that p has a neighborhood that is entirely in A. Hence, p is an interior point of that set A. Hence A is open.
     
  10. Jan 3, 2013 #9

    Erland

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    This doesn't make sense. It must be ##p\notin \overline E##, where ##\overline E## is the closure of ##E##, instead of ##p\notin E##.
    Then, everthing is clear.
     
    Last edited: Jan 3, 2013
  11. Jan 3, 2013 #10
    Since ##p\notin \overline E## then it is not a limit point, hence not every neighborhood of ##p## contains a point of ##E##.

    Let ##N(p)## be the neighborhood with no common points with ##E##.

    What about ##\overline E##? Is the ##\overline E \cap N(p)## an empty set because if it wasn't, then ##N(p)## will contain a limit point of ##E## and these will have neighborhoods that contain a point of ##E##?

    "I understand everything about the proof, except for the part where we go from ##E## to ## \overline E## when we mention the complement. I want to make sure my reasoning is correct" Thanks.
     
    Last edited: Jan 3, 2013
  12. Jan 3, 2013 #11

    Erland

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    You are right, I, and Rudin it seems, were a little bit too quick here. But it is as you write. If ##N(p)## intersects ##\overline E## in a point ##q##, say, then ##N(p)## is also a neighborhood of ##q##, and it must contain an element of ##E##, which was not the case. Thus ##N(p)## is a neighborhood of ##p## which does not intersect ##\overline E##.
     
    Last edited: Jan 3, 2013
  13. Jan 3, 2013 #12
    Thank you. :)
     
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