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I'm having trouble understanding what the $A$ stand for. That's the only place $A$ is mentioned, without knowing what it is, I can't solve the rest of the exercise.

$$V = \{ X \in P(A) \mid |X| \text{ is odd}\},$$

Thanks for your help

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I have no idea, since we have even less context than you do. Could you tell us what exactly the exercise is, and what you've done so far? –  you Mar 25 '12 at 15:12

1 Answer 1

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Suppose that $A$ is a set, then $P(A)$ is the collection of all subsets of $A$. This means that $V$ in your question is the collection of all subsets of $A$ that has an odd number of elements.

For example, if $A$ is the set $\{0,1,2\}$ then $V=\{\{0\},\{1\},\{2\},\{0,1,2\}\}$.

Without further context it is hard to tell what $A$ means exactly, though. I suppose it is a general set on which you are supposed to prove some statement.

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Well, yes, I need to prove that V is a vector space over the field Z2, addition is defined as symmetric difference. Also, is the empty set {} also included in V? The cardinality of {} should be 0, which is an even number, so I would say no? –  gq3 Mar 25 '12 at 16:34
    
@gq3: I suppose that $A$ is given to be any nonempty set, then. Also, the empty set is indeed of even cardinality and therefore not included in $V$. –  Asaf Karagila Mar 25 '12 at 16:36
    
Thanks for your help, turns out it doesn't matter what A actually is, as long it doesn't contain the empty set I can prove it's not a vector space. (Multipling X ∈ V by scalar 0̅ ∈ Z₂ yields an empty set, which is not part of V) –  gq3 Mar 25 '12 at 21:28

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