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Set with a vector space an a group

  1. Dec 4, 2011 #1
    I was thinking this: If I have the set [itex] A = \{ \mathbb{R}^n, \mathbb{R}^n \} [/itex] where for the "first" element I mean the real vector space [itex] \mathbb{R}^n [/itex], and the "second" element is the additive group [itex] \mathbb{R}^n [/itex], then does the set [itex] A [/itex] contain one element ([itex] \mathbb{R}^n [/itex])? Or it contains two elements (a vector space and a group)?

    Maybe this has something to do with category theory, where one considers not only the sets but the structure defined on them. But as sets both have the same elements, so I am confused if it has two elements or just one.

  2. jcsd
  3. Dec 4, 2011 #2


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    Hey Damidami.

    When you use the parentheses, do you mean that A contains the union of R^n vector space and R^n additive group, and by additive group you mean you are talking about a group with the addition operator and the set being R^n?
  4. Dec 4, 2011 #3
  5. Dec 4, 2011 #4


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    I could be wrong, but my understanding is that a group contains a different structure to a vector space.

    A vector space contains your set (in this case R^n) as well as an addition and scalar multiplication operator as well as your standard vector space rules that need to be followed. Lets assume your scalars in scalar multiplication are also real numbers and that addition is standard addition.

    The group object however has a product definition (in this case its just commutative addition) as well as the associated set it deals with (R^n) along with the identity and collections of pairs of elements that are inverses to each other.

    So in regard the above, you have to be careful about what you are considering "the same". The vector space has scalar multiplication whereas the group does not. Also you have the differences about how a group and a vector space are defined in terms of the actual set definitions. I am not 100% sure on the actual set definitions for both the vector space and group objects, but I do know they differ and to answer your question you need to get some definitions for these two objects in terms of their set representation.
  6. Dec 5, 2011 #5
    I agree with chiro. In most books I've read, the authors rigorously define a group as an ordered pair [itex] (G, \cdot)[/itex] of a set and a binary operation such that the group axioms hold. Usually we just refer to G as the group, but really this is an abuse of notation as the operation is just as important as the underlying set.

    Similarly, a vector space is an ordered 4-tuple [itex] (V,F,+,\cdot)[/itex] of the set of vectors (which form an abelian group), the field of scalars, the binary operation of vector addition, and the action of scalar multiplication of F on V.

    So really you can't just say, "I want this [itex] \mathbb{R}^n [/itex] to be a group and this [itex] \mathbb{R}^n[/itex] to be a vector space." [itex] \mathbb{R}^n [/itex] is nothing but a set, with no built-in structure. You are right that we could probably apply a forgetful functor or something to [itex](\mathbb{R}^n, \mathbb{R}, +, \cdot)[/itex] the vector space and get [itex] (\mathbb{R}^n, +)[/itex] the group, but I don't know very much about category theory.
  7. Dec 6, 2011 #6
    Thanks for your answers.
    So the conclusion is the original set I mentioned has two elements, a vector space and a group (with an abuse of notation), and more correctly would be notated as:

    [itex] A = \{ (\mathbb{R}^n, \mathbb{R},+, \cdot), (\mathbb{R}^n, +) \} [/itex]
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