Re: abundance of irrationals!)

Virgil said:
> In article <MPG.1ced35b536936f31989c2b@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > Randy Poe said:
> > >
> > > aeo6 Tony Orlow wrote:
> > > > Randy Poe said:
> > > > > You asked for a proof that a finite set has a largest element.
> > > > >
> > > > > > You started.
> > > > >
> > > > > You didn't see the fifty or so lines after the definition?
> > > > >
> > > > > Let me summarize. I presented an algorithm which:
> > > > > a) is guaranteed to finish and
> > > > > b) at every step, gives you the largest element
> > > > > found so far.
> > > > >
> > > > > The combination of these completes a proof that the
> > > > > algorithm produces a largest element, which establishes
> > > > > that there is a largest element.
> > > > And that proof extends to all of N?
> > >
> > > Nope. Both a) and b) are needed. a) fails for N.
> > Okay, so you cannot find the largest element of N, because you cannot count
> > forever. I thought we kind of established that.
> > >
> > > > Wonderful. N has a largest element. Oh but
> > > > you defined the finite set using its largest element,
> > >
> > > No, I defined the finite set as having n elements. I
> > > didn't "define it using its largest element". I don't
> > > know what the largest element is when I label them as
> > > {a_1, a_2, ..., a_n}. It could be a_17. What makes you
> > > think I'm "defining the set using its largest element"?
> > Okay, you defined the size specifically, which gives you a finite number of
> > iterations to complete the process.
>
> That is all that is needed for a finite set.
That's one way of getting one, like proper subset is one kind of smaller set.
> > >
> ........
>
> > > > Well, I was referring to "definite" sets, as you agreed to call them.
> > > You want
> > > > to reserve "finite" for sets with largest members?
> > >
> > > I want to reserve "finite" for sets whose members can be
> > > numbered from 1 to a natural number n.
> > Same thing. Okay.
>
> Not okay. {} is a finite set without a largest member. By TO's
> definition, it must be an infinite set. Similarly for {"a","b","c"}.
No, it would be an indefinite set. That's the term I used for your infinite set
of finite naturals.
> > >
> > > > Fine. We play the name game.
> > > > The set of all finite naturals is not infinite, and not finite by
> > > your
> > > > definition. Let's say it's "indefinite", just like its largest
> > > member. It's
> > > > certainly not infinite.
>
> "Infinite" means no more thatn "not finite" so that TO now wants to
> create a category of sets that are not finite and not not finite at the
> same time? Does the cantegory of not not not finite follow, and so on ad
> infinitum (or "ad indefinitum" in Orlow-ese?)?
You say it's infinite, I say that's impossible given your restriction on the
element members, which you say are finite, but without bound. Sounds indefinite
to me. It's poorly defined.
>
> > > If it stops, call it "finite".
> > > If it doesn't stop, call it "infinite".
> > > Is there a third possibility?
>
> > Well, yes.
>
> Note: TO revokes and replaces the law of the excluded middle.
> It will, no doubt, be labeled the law of the included muddle.
> als.
> > >
> > > But there exists a way of lining them [rationals]up so that
> > > every rational is labelled with a natural
> > > number. Give me any natural number and I will
> > > tell you what rational is labelled with that value,
> > > what the value of f(n) is.
>
>
> > Yes, that violates my system.
>
> TO's system is irrelevant if it prevents reordering sets to suit one's
> aims.
Yes, and the rules of poker are irrelevant if they prevent you from sneaking in
aces from your pocket when you wish.
>
> > It's a non-quantitative reordering
>
> So what? If it establishes the desired bijection, it is acceptable.
> No set of more than one element has any required ordering AS A SET.
> Orderings, if any, are extra add-ons which may be disacrded whenever
> inconvenient.
It may be acceptable to cardinality, but so are the erroneous conclusions it
produces. That's because it concentrates more on convenience than consistency,
because this permission to discard whatever is inconvenient is the ONLY thing
that allows anyone to draw ANY conclusions from this system. It's like your
concvenient distinction between the size of the set of naturals and the range
of its values, or the convenient discarding of any other logic that might be
brought to bear on the questions at hand, like information theory or infinite
series.

Order is inconvenient to numbers when you want to call them numbers but treat
them like they are not numbers.

>
> The definition of a set, in mathematics dependes only on what are or are
> not members of that set, and in no way implies any inevitable or
> required ordering to that set.
>
> As sets, {1,2,3} = {1,3,2} = {2,1,3} = {2,3,1} = {3,1,2} = {3,2,1}
In most cases, yes. In the case of infinite sets of numbers, the entire set can
only be defined as a function, in which case the values of the elements are
crucial.
>
> > >
> > > > > Can you cite where I made such a statement?
> > > (that finite sets by definition have a largest member)
> > >
> > > > I can't cite where you proved it on the basis of anything but the
> > > definition
> > > > you used.
> > >
> > > OK. So you agree that the proof uses the definition I
> > > used, which does not refer to "largest member". Therefore
> > > you agree that I have never claimed that a finite set
> > > has a largest member "by definition".
> > You deifned a finite set as one which has a definite number of elements. Let
> > me
> > put it that way.
>
> Definition of an "initial segment of N":
> An "initial segment of N" is the set of all members of N less that or
> equal to some n in N under the total linear order induced by the
> successor relation on N.
>
> We shall denote the initial segment deterimed by n as N_n.
>
> For example N_3 = {1,2,3} = {1,3,2} = {2,1,3}
> = {2,3,1} = {3,1,2} = {3,2,1}.
>
Finite initial segment, you mean.

> Definitions of "finite" versus "infinite":
> An arbitrary set, S, is "finite" if and only if there is some n in N for
> which there is a bijection between S and N_n. Otherwise, S is infinite.

No bijections, please, without specifics on the function. Not all bijections
are valid.

>
> Note that, by this definition, each N_n is finite but that N is
> infinite.

Yes, it's poorly defined. Shall I repeat that again?

>
> Note also: an equivalent definition for finite and infinite is:
> S is finite if there does NOT exist any injection from S to any proper
> subset of S, and is infinite if there does exist some such injection
> from S to some proper subset of S.
> NOTE: a function f:A -> B is injective if
> whenever f(x) = f(y) then also x = y.
> (distinct elements have distinct images)
>
>
>
> > If that's the way you define a finite set, then your
> > naturals
> > are not a finite set.
>
> Right!
>
> > But if the elements are all finite, then the set cannot be infinite
>
>
> The way we define infinite, it can be and is.
It's the way you define finite that is at issue, and the generalizations you
make about bijections that are not universal.
>
>
> > > No you didn't, but you have now. Fine. By the way,
> > > "mapping function" is completely general. It doesn't have
> > > to be easily expressible in one line. Any set we
> > > call countable has such a mapping function.
> > Do those multiline functions tend to have inverse functions?
>
> Bijections have inverse functions, non-bijections do not.
>

Then, where there is a bijection, use the function to calculate relative set
sizes. Is that so hard?

--
Smiles,

Tony
.

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