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Difference of closures is subset of closure of difference

  1. Dec 30, 2012 #1
    I can't find the source of this statement now, but I've been trying to prove that

    [tex]\overline{A}\setminus\overline{B} \subseteq \overline{A\setminus B}.[/tex]

    Now [itex]x\in\overline{A}\setminus\overline{B}[/itex] means x is in every closed superset of A but there exists a closed superset of B that doesn't contain x, whereas [itex]x\in \overline{A\setminus B}[/itex] means x is in every closed set that contains every point of A that's not also in B.

    I've tried various manipulations involving equivalent definitions of closure, but have yet to find any obvious way to proceed. Any hints?
     
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  3. Dec 30, 2012 #2

    dextercioby

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    The particular case [itex] A\cap B = \varnothing [/itex] is trivial to prove. If x in A but not in B, then it's in A bar and might be in B bar. Let's assume it's not in B bar too, because the LHS would be then the empty set. So if x is in A bar and not in B bar, then x may be in A or not. If it's in A and not in B bar, then x is not in B, too, so that x would be in A and not in B, hence it would automatically be in (A minus B) bar. The only issue is that x is not in A, but is in A bar. So this x cannot be in B bar, hence cannot be in B, thus cannot be in A minus B but it should be in the closure of A minus B, since the closure of A minus B contains the elements of A bar which are not in B (as was our x).
     
    Last edited: Dec 30, 2012
  4. Dec 30, 2012 #3
    Okay, thanks Dexter! Special case: If [itex]A\cap B = \varnothing[/itex], then [itex]A\setminus B = A[/itex] so [itex]\overline{A\setminus B} = \overline{A}[/itex], and [itex]\overline{A}\setminus\overline{B}\subseteq \overline{A}[/itex], so [itex]\overline{A}\setminus\overline{B}\subseteq \overline{A\setminus B}[/itex].

    General case: [itex]A = (A\setminus B)\cup (A\cap B)[/itex] so [itex]\overline{A} = (\overline{A\setminus B})\cup (\overline{A\cap B})[/itex].

    [tex]\overline{A}\setminus\overline{B}[/tex]

    [tex]=((\overline{A\setminus B})\cup (\overline{A\cap B}))\setminus ((\overline{B\setminus A})\cup (\overline{A\cap B}))[/tex]

    [tex]=(((\overline{A\setminus B})\cup (\overline{A\cap B}))\setminus (\overline{B\setminus A}))\cap (((\overline{A\setminus B})\cup (\overline{A\cap B})) \setminus(\overline{A\cap B}))[/tex]

    [tex]=(\overline{A}\setminus (\overline{B\setminus A}))\cap(\overline{A\setminus B})[/tex]

    [tex]\subseteq \overline{A\setminus B}. \enspace \blacksquare[/tex]
     
  5. Dec 30, 2012 #4
    I didn't follow your meaning here.
     
  6. Dec 30, 2012 #5

    dextercioby

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    The x was chosen to be in A bar but not in B so it could have been in B bar thus the LHS would have been the empty set, due to the arbitrariness of x.
     
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