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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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Michael F. Stemper

Posts: 108
Registered: 9/5/13
Re: Problem from Willard's _General Topology_
Posted: Jan 10, 2014 11:35 AM
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On 01/05/2014 11:55 AM, Brian M. Scott wrote:
> 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:


Hi, Brian! Thanks for taking the time to respond. I'm sorry about
the delay in replying, but I've been chewing over your response,
and I see what you've done. However, I'd still appreciate your
comments on my solution to Part 1. I've broken it down a little
bit more, and switched to your notation.

I'll say "<x_1, ..., x_n>" instead of saying "P_n". In that case, my
response to Part 1 is:

<x_1> = { {x_1} }

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

Letting n=1 gives <x_1, x_2> =
<x_1> U { (U <x_1>) U {x_2} } =
{ { {x_1} } U { (U { {x_1} }) U {x_2} } =
{ { {x_1} } U { {x_1} U {x_2} } } =
{ { {x_1} } U { {x_1, x_2} } } =
{ {x_1}, {x_1, x_2} }

This is the same definition of <x_1, x_2> that Willard gave in 1C.

Letting n=2 gives <x_1, x_2, x_3> =
<x_2> U { (U <x_2>) U {x_3} } =
{ {x_1}, {x_1, x_2} } U { (U { {x_1}, {x_1, x_2} }) U {x_3} } =
{ {x_1}, {x_1, x_2} } U { {x_1, x_2} U {x_3} } =
{ {x_1}, {x_1, x_2} } U { {x_1, x_2, x_3 } } =
{ {x_1}, {x_1, x_2}, {x_1, x_2, x_3 } }

And, in general, <x_1, ..., x_n> =
{ {x_1}, ..., {x_1, ..., x_n} }

Is this a valid definition of "ordered n-tuple"?

--
Michael F. Stemper
I feel more like I do now than I did when I came in.



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