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Topic: Rudin and nonsense
Replies: 29   Last Post: May 3, 2012 9:46 PM

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 Paul A. Tanner III Posts: 5,920 Registered: 12/6/04
Re: Rudin and nonsense
Posted: Apr 29, 2012 12:55 PM

On Sun, Apr 29, 2012 at 4:05 AM, Didacticus <ddidacticus@yahoo.com> wrote:
> Nothing I wrote suggested Rudin's choice was *arbitrary* -- I simply pointed out it was a *choice*.

Well, then, the entire textbook that Rudin wrote is the result of choice. And so your talk of "choice" means nothing.

And this "choice" gives us Theorem 1.29, the *equality* - and not mere equivalence - (a,b) = a + bi.

>
> The nonsense I alluded to is you going in circles ...
>
> ... and then accusing people of "denying" Rudin when they dare to point out the silliness.

Why do you think Rudin is going in circles?

If I'm going in circles, then so is Rudin, since all I'm doing is laying out in detail what Rudin did, where the sequence of *equalities* - and not mere equivalences -

(a,0) = a + 0i = a + 0 = a

and so

(a,0) = a

is the end of what Rudin did with definitions and theorems 1.26 through 1.29.

The equality (a,0) = a that follows from Theorem 1.29 is so trivially and clearly true as an application of Theorem 1.29, Rudin did not of course bother to actually write it out - he left that to us. And so the equality (a,0) = a may be viewed as simply an unwritten corollary to Theorem 1.29.

You need to process what I in my prior post

"Re: Rudin and nonsense"
http://mathforum.org/kb/message.jspa?messageID=7807322

actually wrote. Here is the most relevant part that proves that your thinking that Rudin is going in circles is pure silliness:

Quote: "That is, taken as a whole, Rudin proved isomorphism in Theorem 1.26 by which (a,0) and a for all real a are at least equivalent (note that I did not say "equal"), but then by replacement and by Theorem 1.29, we find that not only are (a,0) and a equivalent, but that they are equal as well."

Your thinking that Rudin is going in circles results from your mistake in thinking that equivalence is equality. It is not. Isomorphism is an equivalence relation. By definition, equivalence is necessarily a relation that is reflexive, symmetric, and transitive but it not necessarily antisymmetric, while by definition equality is necessarily all four.