From: Alan Smaill on
Lester Zick <dontbother(a)nowhere.net> writes:

> On Sun, 15 Apr 2007 21:34:09 +0100, Alan Smaill
> <smaill(a)SPAMinf.ed.ac.uk> wrote:
>
>>>>>>>>>>Dear me ... L'Hospital's rule is invalid.
>>>
>>> So returning to the original point, would you care to explain your
>>> claim that L'Hospital's rule is invalid?
>>
>>haha!
>
> Why am I not surprized? Remarkable how many mathematiker opinions on
> the subject of mathematics don't quite hold up to critical scrutiny.

knew you wouldn't get it,
irony is not a strong point with Ziko.

nor indeed do you bother defending your own view that you can use
Hospital to work out the value for 0/0.

well, there you go.


> ~v~~

--
Alan Smaill
From: K_h on

"Virgil" <virgil(a)comcast.net> wrote in message
news:virgil-295022.17233815042007(a)comcast.dca.giganews.com...
> In article <87WdnbF6VIxxMr_bnZ2dnUVZ_gqdnZ2d(a)comcast.com>,
> "K_h" <KHolmes(a)SX729.com> wrote:
>
>> "Virgil" <virgil(a)comcast.net> wrote in message
>> news:virgil-19C008.15430215042007(a)comcast.dca.giganews.com...
>> > In article <VZGdnTTzjvOPG7_bnZ2dnUVZ_hGdnZ2d(a)comcast.com>,
>> > "K_h" <KHolmes(a)SX729.com> wrote:
>> >
>> >> "Virgil" <virgil(a)comcast.net> wrote in message
>> >> news:virgil-B82679.22384714042007(a)comcast.dca.giganews.com...
>> >> > In article <qbOdnddPifcj97zbnZ2dnUVZ_rylnZ2d(a)comcast.com>,
>> >> > "K_h" <KHolmes(a)SX729.com> wrote:
>> >> >
>> >> >> "Virgil" <virgil(a)comcast.net> wrote in message
>> >> >> news:virgil-9A74B0.17144214042007(a)comcast.dca.giganews.com...
>> >> >> > In article <JvmdneShNPwdxLzbnZ2dnUVZ_tCtnZ2d(a)comcast.com>,
>> >> >> > "K_h" <KHolmes(a)SX729.com> wrote:
>> >> >> >
>> >> >> >
>>
>> <removed for size>
>>
>> >> I did not mean equality as in the same set. From Wikipedia, a set is
>> >> ordered if every pair of its members is mutually comparable. If, for
>> >> any
>> >> two distinct members x y, we cannot tell if any of these are true
>> >> (x<y,
>> >> x=y,
>> >> y<x) then the set is not ordered. Consider a set of four people and
>> >> the
>> >> set
>> >> is ordered by height. Sam if 5 feet tall, Jane is 6 feet tall, John
>> >> is 6
>> >> feet tall, and Mike is 7 feet tall. The set {Sam, Jane, John, Mike}
>> >> is
>> >> ordered by:
>> >>
>> >> Sam < Jane = John < Mike
>> >>
>> >> Clearly every pair of members of this set is comparable and Jane =
>> >> John
>> >> under this ordering. Equality here means that Jane and John are
>> >> comparable
>> >> and equal under the ordering; it does not mean that Jane is John.
>
> Then the set is not totally ordered, but only partially ordered.
> In a totally and strictly ordered set of any two different members, one
> MUST be larger and the other smaller.

I am still looking for an authoritative source for this. The example set I
gave was totally ordered, in the sense that every member of the set is
mutually comparable, but, just to make it absolutely clear, I will enumerate
the entire order:

Sam < Jane
Sam < John
Sam < Mike
Jane = John
Jane < Mike
John < Mike

This total order satisfies the law of trichotomy: For any x and y in the
ordered set, either (x >= y) or (y >= x).

> This
>> >> is
>> >> an ordered set by Wikipedia's definition because all of its members
>> >> are
>> >> mutually comparable under the relation. Are you claiming that an
>> >> ordered
>> >> set requires x<y or y<x for any two distinct members x and y?
>> >
>> > A totally ordered set does.
>>
>> In mathworld's definition for a totally ordered set, I still don't see
>> where
>> there is a requirement that x<y or y<x for any two distinct members.
>
> Trichotomy!
> <quote>
> 4. Comparability (trichotomy law): For any a and x , either x >= y or
> y >= x.
> <unquote>
>
> This means that whenever x = y does NOT hold then either x>y or y>x.

Sam < Jane
Sam < John
Sam < Mike
Jane = John
Jane < Mike
John < Mike

In the above ordered set, x=y does not hold for the below cases:

Sam < Jane
Sam < John
Sam < Mike
Jane < Mike
John < Mike

These cases satisfy x>y or y>x.

>> Can you provide a source where the definition of a totally ordered set
>> *requires* that x<y or y<x for any two distinct members x and y? If so
>> then
>> mathworld's definition for a totally ordered set is ambiguous.
>
> See above! Unless x = y, one must have either x<y or y<x !!!!

You seem to be saying that, for totally ordered sets, equality only exists
in the case of identity and therefore equality among two distinct members
cannot be defined in the order. If that is the case then what you are
saying is correct. So what I would like is an authoritative source that
says that. Both Wolfram and Wikipedia do not say that. If they did then
the definition would say something like "...for any distinct members x y,
x<y or y<x" and there wouldn't be statements like "...for any x and y, x<y
or y<x or y=x".

I am totally prepared to admit that your understanding is correct and both
Wolfram and Wikipedia are badly worded. There is lots of misinformation on
the internet. What troubles me is that the books on set theory I have also
have their definitions worded like wolfram and Wikipedia. All I am asking
from you is an authoritative source whose definition of a totally ordered
set clearly states that equality among two distinct members cannot be
defined in the order.


K_h


From: Michael Press on
In article
<1176500762.035139.250710(a)n59g2000hsh.googlegroups.com>,
"MoeBlee" <jazzmobe(a)hotmail.com> wrote:

> On Apr 13, 12:11 pm, Tony Orlow <t...(a)lightlink.com> wrote:
> > I had said that, hoping you might give some explanation, but you didn't
> > really.
>
> Since you posted that, I wrote a long post about the axiom of choice.
> Now it's not showing up in the list of posts. Darn! I went into a lot
> of detail and answered your questions; I don't want to write it all
> again; maybe it will show up delayed.

* Composer your messages in a local file.

* Set a preference in your news reader to write to a local file
a copy, complete with headers, of all your posted messages.


* Set a preference in your news reader to mail to a specified
address, a copy, complete with headers, of all your posted
messages.

I am surprised I have to explain this.

--
Michael Press
From: Lester Zick on
On Mon, 16 Apr 2007 01:03:22 +0100, Alan Smaill
<smaill(a)SPAMinf.ed.ac.uk> wrote:

>Lester Zick <dontbother(a)nowhere.net> writes:
>
>> On Sun, 15 Apr 2007 21:34:09 +0100, Alan Smaill
>> <smaill(a)SPAMinf.ed.ac.uk> wrote:
>>
>>>>>>>>>>>Dear me ... L'Hospital's rule is invalid.
>>>>
>>>> So returning to the original point, would you care to explain your
>>>> claim that L'Hospital's rule is invalid?
>>>
>>>haha!
>>
>> Why am I not surprized? Remarkable how many mathematiker opinions on
>> the subject of mathematics don't quite hold up to critical scrutiny.
>
>knew you wouldn't get it,
>irony is not a strong point with Ziko.

I'll grant you I'm much more adept at hyperbolic rhetorical irony than
simplistic irony which tends to go right past me much as mathematitcs
tends to go right past you.

>nor indeed do you bother defending your own view that you can use
>Hospital to work out the value for 0/0.

I never claimed L'Hospital's rule was valid. I assume its validity or
at least its utility was established by L'Hospital much as I assume
the validity or at least the uitility of 1+1=2 has been established by
others just as I assume you're a mathematiker because you're too lazy
or stupid to demonstrate the truth of your arguments but not too lazy
or stupid to formulate arguments whose truth you can't demonstrate.

>well, there you go.

Yes indeedy do, Snail.

~v~~
From: MoeBlee on
On Apr 14, 7:08 pm, Tony Orlow <t...(a)lightlink.com> wrote:

> Maybe I don't understand Robinson as well as I should, but it seems to
> me the basis of his analysis was semantic, regarding statements that
> would be considered true of *N if true of N. But, do go on...

"But do go on..." Hey, smartass, YOU'RE the one who waves non-standard
analysis like banner when you don't know ANYTIHNG about it.

If you want to know anything about non-standard analysis, then you'd
start by learning basic predicate calculus, set theory, and
mathematical logic (personally, I'd recommend that order).

> > The MAJOR point - the hypothetical nature of mathematical reasoning
> > (think about the word 'if' twice in the poster's paragraph) and the
> > inessentiality of what words we use to name mathematical objects and
> > their properties.
>
> I guess, by "inessentiality", you mean any attribute that one could
> assign to any object... IF that's what you mean, or not....

No, smartass, months ago, several times, I explained in detail to you,
and with respect to the technicalities, what I mean.

> > I've been trying to get you to understand that for about two years
> > now.
>
> Perhaps what you mean is exactly what I am trying to get across to
> Lester. If the truth table is the same, then it's the same logical
> function. It doesn't matter what the parameters are, it always works the
> same way. The pattern defines the relationship.

That's getting closer. I won't quibble with it. But I'll just say that
in mathematical logic we have even more precise ways of saying it.

> On the subject of ifs, "if this then that" means logical implication, or
> causality, and sometimes it's hard to tell which is meant, or if they're
> being confused. Is that what you "mean"?

No.

> >> I don't take transfinite cardinality to mean
> >> "size". You say I missed the point. You didn't intersect the line.
>
> > You just did it AGAIN. We and the poster to whom you responded KNOW
> > that you don't take cardinality as capturing your notion of size. The
> > point is then just for your to recognize that IF by 'size' we mean
> > cardinality, then certain sentences follow and certain sentences don't
> > follow and that what is important is not whether we use 'size' or
> > 'cardinality' or whatever word but rather the mathematical relations
> > that are studied even if we were to use the words 'schmize' or
> > 'shmardinal' or whatever.

THAT, to which you did not respond, is, at least at this informal
level of discussion with you, good enough for a start as to what I
mean.

MoeBlee