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Topic: Problem from Willard's _General Topology_
Replies: 8   Last Post: Jan 11, 2014 8:54 PM

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Brian M. Scott

Posts: 1,289
Registered: 12/6/04
Re: Problem from Willard's _General Topology_
Posted: Jan 5, 2014 12:55 PM
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On Sun, 05 Jan 2014 11:09:10 -0600, "Michael F. Stemper"
<michael.stemper@gmail.com> wrote in
<news:lac3jp$kfd$1@dont-email.me> in alt.math.undergrad:

> I've started on _General Topology_ by Stephen Willard, and
> am having a little difficulty with Problem 1D, "Cartesian
> Products".


> Part 1 of this problem reads:
> Provide an inductive definition of "the ordered n-tuple (x_1, ...,
> x_n) of elements x_1, ..., x_n of a set" so that (x_1, ..., x_n) and
> (y_1, ..., y_n) are equal iff their coordinates are equal in order,
> i.e., iff x_1=y_1, ..., x_n=y_n.


I prefer to use angle brackets for ordered pairs and ordered
n-tuples generally and will do so here. You need to have
done 1C first, so that you have <x,y> = {{x},{x,y}} as
your base case (n=2). Then given <x_1, ..., x_n> and an
element x_{n+1}, define

<x_1, ..., x_{n+1}> = <<x_1, ..., x_n>, x_{n+1}>.

If you prefer, you can associate to the right: given
<x_1, ..., x_n> and an element x_0, define

<x_0, ..., x_n> =<x_0, <x_1, ..., x_n>>.

The first option means that <x_1, x_2, x_3>, for instance,
is actually <<x_1, x_2>, x_3>; the second makes it
<x_1, <x_2, x_3>> instead.

> My response to this makes use of the shorthand notation that
> P_n = (x_1, ..., x_n). For instance, P_3 = (x_1, x_2, x_3).


Notationally this doesn?t make much sense: you?re trying to
write a recursive construction as if it were a proof by
induction. You don?t have an object P(n) for each n; you
simply have the definition of ?ordered n-tuple? for each n.

[...]

> (b) seems simple enough, as well. Define the trivial
> Cartesian product of X_1 as { {x} | x in X_1 }. Then, let
> the Cartesian product X_1 * ... * X_n * X_(n+1) be
> defined as the product (X_1 * ... * X_n) * X_(n+1).


Yes, this will work, though it would be better to write

(X_1 x X_2 x ... x X_n) x X_{n+1}

if you don?t have easy access to

(X_1 × X_2 × ... × X_n) × X_{n+1}.

This matches the first of my two approaches to defining
ordered n-tuples. You could also associate the other way,
matching the second.

[...]

Brian



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