Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
  2. jcsd
  3. Dec 30, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    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 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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook