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Cl(A) = A' ?

  1. Apr 16, 2012 #1
    Cl(A) = A' ??

    Unraveling the definitions, I keep getting that Cl(A) = A'.

    [tex]x\in \overline{A}[/tex]

    [tex]\Leftrightarrow (\forall U\in \tau)[(A\subseteq X\setminus U)\Rightarrow (x\in X\setminus U)][/tex]

    [tex]\Leftrightarrow (\forall U\in \tau)[\neg (x\in X\setminus U)\Rightarrow \neg(A\subseteq X\setminus U) ][/tex]

    [tex]\Leftrightarrow (\forall U\in \tau)[(x\in U)\Rightarrow (A \cap U \neq \emptyset ) ][/tex]

    [tex]\Leftrightarrow x\in A'.[/tex]

    (The empty set is its own closure, so if x is in A, then A is not empty.)

    I suspect the problem may lie in the substitution

    [tex]\neg (x\in X\setminus U) \Leftrightarrow \neg((x\in X)\& \neg(x\in U))[/tex]

    [tex]\Leftrightarrow \neg(x\in X) \vee (x\in U)[/tex]

    [tex]\Leftrightarrow (\forall x\in X) [x\in U].[/tex]

    On it's own, the final step of deleting this "for all X" looks sound to me (we're already implicitly talking about all x in X, so why do we need to consider the possibility that x is not in X?), but in the above context, I've moved from "for all x in X, if P is true or x is in X, then ..." (which is true of all x in X) to "for all x in X, if P is true ..." (which is not necessarily true of all x in X).

    Can anyone help me understand what the logical rule is here? (I.e. if this isn't a legitimate substitution, what general rule makes it illegitimate.) Is this why I'm getting the anomalous result that Cl(A) = A'?
     
    Last edited: Apr 16, 2012
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  3. Apr 16, 2012 #2

    micromass

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    Re: Cl(A) = A' ??

    What is the definition of A' ??
     
  4. Apr 16, 2012 #3
    Re: Cl(A) = A' ??

    Ooh, ooh, I see the mistake! I was garbling the definition of A':

    [tex]\left \{ x | (\forall U)[(x\in U)\Rightarrow (U\setminus \left \{ x \right \} \cap A \neq \emptyset)] \right \},[/tex]

    rather than simply

    [tex]\left \{ x | (\forall U)[(x\in U)\Rightarrow (U \cap A \neq \emptyset)] \right \}.[/tex]

    So actually [itex]\overline{A}=A\cup A'[/itex]. If x is in the closure of A, either x is a limit point of A, or x belongs to A.
     
  5. Apr 16, 2012 #4

    micromass

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    Re: Cl(A) = A' ??

    Aah yes, I found your A' a bit weird in the OP :smile: Good you found the mistake!
     
  6. Apr 16, 2012 #5
    Re: Cl(A) = A' ??

    It dawned on me just before I read your hint! Thanks, micromass - ever ready to spring to my rescue : )

    Did my question about the substitution make sense?
     
  7. Apr 16, 2012 #6

    micromass

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    Re: Cl(A) = A' ??

    Yeah. Your substitution looks ok to me.
     
  8. Apr 16, 2012 #7
    Re: Cl(A) = A' ??

    And I see why now, at last! The statement is actually of the form [itex]P\vee Q \Rightarrow R[/itex], where P is not true. That being the case, the antecedent is equivalent to Q. I think it was the double negation that confused me.
     
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