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I'm stuck trying to figure out how to prove that $(\mathcal P (\mathbb N) \space , \space \triangle)$ is an abelian group? I know the definition, but I'm confused how to incorporate the fact that I'm dealing with a powerset instead of just the integers or natural numbers, so I'm seeking help to understand how proofs on sets works.

Also, I'm using $\triangle$ to mean symmetric difference.

Martin Sleziak
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Benzne_O
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Possibly the best approach is to note that if $A$ is an arbitrary set, and you associate to each element of $B \in \mathcal P ( A)$ its characteristic function $$ \textbf{1}_{B} : {A} \to \mathbf{Z}_{2}, f(a) = \begin{cases} 1 & \text{if $a \in B$},\\ 0 & \text{if $a \notin B$,}\end{cases} $$ then symmetric difference becomes (pointwise) sum of characteristic functions, $$ \textbf{1}_{B} + \textbf{1}_{C} = \textbf{1}_{B \Delta C} $$

Andreas Caranti
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    This is the best, mainly when we verify the associativity.+1 –  Mar 09 '13 at 18:56
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    I agree. On the other hand, associativity is also clear for the sets directly: Both $A \Delta (B \Delta C)$ and $(A \Delta B) \Delta C$ consist of those elements which belong to exactly one of the sets $A,B,C \subseteq X$. – Martin Brandenburg Mar 09 '13 at 19:08
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    @MartinBrandenburg, absolutely, that is, those element that belong to an odd number ($1$ or $3$ here) of the sets $A, B, C$. Still I find instructive how a detour may reduce the problem to a standard, general argument. – Andreas Caranti Mar 09 '13 at 19:26
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Just check the axioms. Prove that symmetric difference is commutative (which is clear from its definition) and associative (which takes a little work). Prove that there’s an identity, i.e., a set $Z$ such that $Z\triangle A=A$ for all $A\in\wp(\Bbb N)$; there’s exactly one set that has that property, and it’s not hard to find. Finally, check that every $A\in\wp(\Bbb N)$ has an inverse. A hint for both of those last parts: what’s $A\triangle A$?

Brian M. Scott
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  • Not quite sure about the hint, but would the empty set work as an identity element? I believe that A $\triangle \space \emptyset$ = A, but I could be wrong – Benzne_O Mar 09 '13 at 21:03
  • @Benzne_O: You’re right: $A\triangle\varnothing=(A\setminus\varnothing)\cup(\varnothing\setminus A)=A\cup\varnothing=A$ for all $A$. Thus, $\varnothing$ works just fine as the identity. – Brian M. Scott Mar 09 '13 at 21:05
  • Alright thanks, this answer made the most sense to me. – Benzne_O Mar 09 '13 at 21:07
  • @Benzne_O: You’re welcome. Feel free to ask for further clarification if you need it. – Brian M. Scott Mar 09 '13 at 21:07
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In fact $(\mathcal P (X) \space , \triangle)$ is a group for any $X$:

  • closure is immediate since $\triangle : \mathcal P (X) \times \mathcal P (X) \to \mathcal P (X)$
  • identity: $x \triangle \{\} = x$
  • inverses: $x \triangle x = \{\}$ so everything is its own inverse.
  • associativity: $(x \triangle y) \triangle z = x \triangle (y \triangle z)$ is obvious from a venn diagram.

and it's abelian because $\triangle$ is symmetric.