Re: Questions about Cartesian product


Arturo Magidin wrote:
> In article <dfjq7u$hvg$1@xxxxxxxxxxxxxxxx>,
> Lee Rudolph <lrudolph@xxxxxxxxx> wrote:
> >magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) writes:
> >
> >...
> >>Given a family {A_i} i in I of sets, the
> >>Cartesian product is by definition the set
> >>
> >>{ f:I -> Union(A_i) : f(i) in A_i for each i}
> >>
> >>in other words, the set of all choice functions.
> >...
> >>>2.- It is intuitively obvious that the product of sets A and B is the
> >>>empty set if, say, A is empty. How can this be proved?
> >>
> >>By definition. In this case, the index set I is I={A,B}, with the "A"
> >>set being A and the "B" set being B.
> >
> >Are you kidding on the square?
>
> Huh?
>
> No, I'm not kidding. If you have a set, the easiest way to consider
> the set as a family of sets is to let the set itself be the index set,
> and letting each index correspond to itself. Given two sets, A and B,
> we know there is a set that contains A and B, hence I={A,B} is a set,
> and then you can consider I as the index set and define the cartesian
> product in the general way, as the set of all choice functions
> f:I->A U B such that f(A) is in A and f(B) is in B.
>

But ... doesn't this run into a small problem when one
comes to defining A x A ?? (I thought that was the point
of the "Rudolph-esque" comment ... ) Looks like you only
get the diagonal of A x A with this approach ...


> --
> ======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
> ======================================================================
>
> Arturo Magidin
> magidin@xxxxxxxxxxxxxxxxx

.

Relevant Pages

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  • Re: Questions about Cartesian product
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