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[parent] properties of symmetric difference (Derivation)

Recall that the symmetric difference of two sets $ A,B$ is the set $ A\cup B-(A\cap B)$. In this entry, we list and prove some of the basic properties of $ \triangle $.

  1. (commutativity of $ \triangle $) $ A\triangle B=B\triangle A$, because $ \cup$ and $ \cap$ are commutative.
  2. If $ A\subseteq B$, then $ A\triangle B=B-A$, because $ A\cup B= B$ and $ A\cap B =A$.
  3. $ A\triangle \varnothing = A$, because $ \varnothing \subseteq A$, and $ A-\varnothing=A$.
  4. $ A\triangle A=\varnothing$, because $ A\subseteq A$ and $ A-A=\varnothing$.
  5. $ A\triangle B=(A-B)\cup (B-A)$ (hence the name symmetric difference).
    Proof. $ A\triangle B= (A\cup B)-(A\cap B)= (A\cup B)\cap (A\cap B)'= (A\cup B)\cap (A'... ...p B)\cap A')\cup ((A\cup B)\cap B')= (B\cap A')\cup (A\cap B')= (B-A)\cup (A-B)$. $ \qedsymbol$
  6. $ A'\triangle B'=A\triangle B$, because $ A'\triangle B'=(A'-B'')\cup (B'-A'')=(A'\cap B)\cup (B'\cap A)=(B-A)\cap (A-B)=A\triangle B$.
  7. (distributivity of $ \cap$ over $ \triangle $) $ A\cap (B\triangle C)=(A\cap B)\triangle (A\cap C)$.
    Proof. $ A\cap (B\triangle C) = A\cap ((B\cup C)-(B\cap C))$, which is $ (A\cap (B\cup C))-(A\cap (B\cap C))$, one of the properties of set difference (see proof here). This in turns is equal to $ ((A\cap B)\cup (A\cap C))-((A\cap B)\cap (A\cap C)) = (A\cap B)\triangle (A\cap C)$. $ \qedsymbol$
  8. (associativity of $ \triangle $) $ (A \triangle B) \triangle C = A \triangle (B \triangle C)$.
    Proof. Let $ U$ be a set containing $ A,B,C$ as subsets (take $ U=A\cup B\cup C$ if necessary). For a given $ B$, let $ f:P(U)\times P(U)\to P(U)$ be a function defined by $ f(A,C)= (A\triangle B)\triangle C$. Associativity of $ \triangle $ is then then same as showing that $ f(A,C)=f(C,A)$, since $ A\triangle (B\triangle C)= (B\triangle C)\triangle A = (C\triangle B)\triangle A$.

    By expanding $ f(A,C)$, we have

    $\displaystyle (A\triangle B)\triangle C$ $\displaystyle =$ $\displaystyle ((A\triangle B)-C)\cup (C-(A\triangle B))$  
      $\displaystyle =$ $\displaystyle (((A-B)\cup (B-A))\cap C')\cup (C- ((A\cup B)-(A\cap B)))$  
      $\displaystyle =$ $\displaystyle (((A\cap B')\cup (B\cap A'))\cap C')\cup ((C\cap A\cap B)\cup (C-(A\cup B))$  
      $\displaystyle =$ $\displaystyle ((A\cap B'\cap C')\cup (B\cap A'\cap C'))\cup ((C\cap A\cap B)\cup (C\cap A'\cap B'))$  
      $\displaystyle =$ $\displaystyle (B\cap A'\cap C')\cup (B\cap A\cap C)\cup (B'\cap A\cap C')\cup (B'\cap A'\cap C).$  

    It is now easy to see that the last expression does not change if one exchanges $ A$ and $ C$. Hence, $ f(A,C)=f(C,A)$ and this shows that $ \triangle $ is associative. $ \qedsymbol$

Remark. All of the properties of $ \triangle $ on sets can be generalized to $ \triangle $ on Boolean algebras.



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Cross-references: Boolean algebras, expression, easy to see, function, necessary, subsets, associativity, proof, properties of set difference, distributivity, commutativity, properties, symmetric difference
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This is version 10 of properties of symmetric difference, born on 2004-09-18, modified 2008-05-01.
Object id is 6192, canonical name is ProofOfTheAssociativityOfTheSymmetricDifferenceOperator.
Accessed 7194 times total.

Classification:
AMS MSC: 03E20 (Mathematical logic and foundations :: Set theory :: Other classical set theory )
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