Re: abundance of irrationals!)

Randy Poe said:
>
> aeo6 Tony Orlow wrote:
> > Randy Poe said:
>
> > > No, that's the definition. Most proofs have steps beyond
> > > the definition. You asked for a definition, so I gave that
> > > first.
> [SNIP]
> > > > > > Axioms, please, and logic. No leaps.
> > >
> > > Did my best to comply, starting with the definition.
> > > Sorry that you took issue with my attempt to begin
> > > where you asked me to begin.
> > You didn't finish. Derive for me the fact that a finite set
> necessarily has a
> > finite number.
>
> 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? Wonderful. N has a largest element. Oh but
you defined the finite set using its largest element, so you kind of proved
that x=x. Very impressive.
>
> > > Another starts from the bijective definition of infinite
> > > sets. Do you have another in mind?
> > Bounded, perhaps? All I know is that what we know about infinite
> series
> > contradicts the notion of an infinite set of distinct finite
> naturals. There is
> > no clear line between finite and infinite, value-wise.
>
> Let me know when you've pinned down a definition of
> "finite set", and then we can discuss the properties.
>
> If you mean bounded sets then no, not all bounded sets
> have a largest member. For instance the set of rationals
> less than sqrt(2) has no largest member.
Sort of like the set of all reals less that sqrt(2)? Hmmm....
>
> > Sure, by "definition". I have no problem with seeing that such sets
> exist and
> > are commonplace, and have largest members if they all have distinct
> quantities
> > associated with them.
>
> OK, good. Then the proof is unnecessary, since you agree that
> the sets I call finite have larges members.
Well, I was referring to "definite" sets, as you agreed to call them. You want
to reserve "finite" for sets with largest members? 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.
>
> Now we just have to figure out what you mean by "finite
> set", and then we can discuss whether N is one of them.
>
> > > Two comments: I defined a finite set as one that has
> > > elements that can be labeled from 1 to n for some natural
> > > number n. There's no particular order implied, so
> > > there's no definite "last element". You can count
> > > them in any order you want.
> > As long as you go from 1 to n. Order is more important in many cases
> than set
> > theory admits.
>
> No, it's exactly as important as I said. The "going from
> 1 to n" is after you apply (arbitrary) labels to the
> elements.
>
> It works just well to count {1,2,3,4,5} in that order
> as in the order 5,1,4,2,3.
>
> Are you claiming the cardinality depends on which
> element I call a_1, which a_2, etc?
It depends on the mapping function and value range.
>
> I WILL claim that your incomplete definition of cardinality
> suffers from this weakness, but not the bijective one.
Which weakness?
>
> > You said finite sets by definition have largest elements,
>
> No I didn't. The word "largest" is not part of the definition.
No, it's not.
>
> I said that finite sets have bijective maps to {1,2,3,...,n}
>
> That does not imply I'm applying those labels in order
> of size of element. I'm claiming that if you give me
> a shuffled deck of cards, then my maximum-finding
> algorithm will find the largest card (you have to
> add an ordering of suits to make it a complete order).
> I'm not saying the highest card has to be the last
> card. I'm just saying that in a finite number of
> steps I can find a highest card.
You also have to define if the suit or number has precedence. You have a 2D
matrix, 4x13=52. Rationals have a 2D matrix as well: NxN=N^2 rationals.
>
>
> > and you offered a definition that doesn't mention anything
> > about largest members,
>
> I offered a *proof* from the *actual definition* that
> finite sets have largest members. If I *said* that the
> definition of finite sets is that they have largest
> members, that was wrong. Can you cite where I made such
> a statement?
I can't cite where you proved it on the basis of anything but the definition
you used.
>
> I'm getting off this loop to talk about the Tony-cardinality
> measure. Let's see your responses...
> > > > The size of a set of real numbers defined using a function f(n)
> on
> > > the naturals
> > > > from x to y,
> > >
> > > What does this mean?
> > >
> > > Suppose x = {1.3, 2.6, 5.7} and y = {0.5, 0.4, sqrt(37)}.
> > >
> > > What is "f(n) on the naturals from x to y"?
> > > Can you give an example? Normally f:x->y, a
> > > mapping from x to y, takes elements of x and
> > > produces elements of y. You are talking about
> > > three sets: x, y and N. I don't know what you have
> > > in mind.
> > Those don't look like the kind of sets I have defined this method as
> > addressing, now do they?
>
> I was reacting to your first statement, that you are
> talking about sets of real numbers. So I assumed you
> were talking about sets of real numbers. If that's not
> the kind of sets you have defined this method for,
> then what are?
Sets defined by mapping functions from the naturals to the reals which have
function inverses. I stated that up front.
>
> Do you want to rephrase this?
> "The size of a set of real numbers defined using a function f(n) on
> the naturals from x to y..."
"y in the reals". You left that out. naturals in, reals out.
>
> It doesn't quite parse and also (in retrospect) doesn't have
> anything in common with what you're defining. For instance,
> this is something new:
>
> > I don't see a formula defining them, do you?
>
> So now you are restricting your sets to sets of... something
> ... for which there is a formula defining them?
and invertible function st f(g(x))=g(f(x))=x.
>
> Wait a sec.
>
> Is this what you wanted to say?
>
> "Let x be a set of real numbers of the form {x:x = f(n) where
> f(n) has the following properties
> - It is defined for every natural number n
> - It is expressible by a "formula"
> - It is invertible
> - ???
> - ???"
>
> By the way, my sets x and y are expressible by formulas,
> as is any set of finite real numbers.
>
> x = {1.3, 2.6, 5.7}
>
> is definable as {f(1), f(2), f(3)} where
>
> f(n) = [1.3*(n-2)(n-3)/2] - [2.6*(n-1)(n-3)] +
> [5.7*(n-1)(n-2)/2]
>
Okay, of course you can create a function that contains some finite set of
points. If you want to boil this down and derive an inverse function, then
apply it to 5.7 and 1.3 and subtract, I am quite confident you will get......3!
Give it a try. Live a little!!!

> I could just as easily express x as {f(0),f(1000),f(6)}.
> And no I didn't know that when I wrote down x. I'm
> just giving you the Lagrange interpolation formulas,
> which are a technique for finding a polynomial
> to pass through any finite set of points.
Yes I got that. I am not sure what you are saying in the beginning of this
paragraph though.
>
> > If I say I can prove something about red apples, do
> > you complain that it doesn't apply to your green bananas?
>
> No, but if you say you can prove something about sets
> of real numbers, I'll complain if it doesn't apply to
> sets of real numbers.
I defined which sets it applies to, up front. Don't blame me for your inability
to read an entire finite set of characters.
>
> > > So f(n) takes elements of N and produces elements of x. There
> > > is no y.
> > Yes a function from the naturals to the reals, as I said.
>
> No. Look at your introductory sentence again. It does not
> say "a function from the naturals to the reals". It says
> "a set of real numbers defined using a function f(n) on
> the naturals from x to y". It says we're talking about
> a set of real numbers, but you didn't like my set of
> real numbers. It talks about two sets x and y without
> saying what kind of sets these are. It talks about f(n)
> without saying what kind of thing f(n) produces.
x and y are variables, of the natural and real variety, respectively. Functions
generally are performed on numbers, and yield numbers, or we usually call them
"operations", I think. F(n) produces a set of reals from the set of naturals.
>
> Since you are saying now that f maps from the naturals
> to the reals (and has some other unstated properties,
> which we'll iron out in discussion), and since my
> x has that property, I fail to see what's wrong with
> it.
Now that uou have derived a funstion defining it, apply the method and see
whether you get 3. If you do, will you at least be satisfied that it has some
validity?
>
> So no, I still don't know what type of sets you are
> restricting yourself to. But if there is ANY restriction,
> then it's not a general theory of set cardinality
> is it?
It enables precise measures of sets which currently are all thrown into the
"countable" bag. It's progress, no?
>
> > If you don't
> > understand those words, it is a function that takes a natural number
> as a
> > parameter, and produces a real number result (which may of course be
> a
> > natural).
>
> Then normally you would say so.
I thought I did.
>
> >
> > When we say f(x)=*&^(&^, that is the same as y=*&^(&^. f(x)=y, when
> you are
> > graphing functions. Where did you learn math?
>
> Beyond high school, inputs to maps are not always numbers, are
> not always called x, outputs are not always called y,
> multiple dimensions are allowed, not everything is
> graphable, etc.
So? Why did you ask where y was? Did you not understand that's what f(x) is?
Don't ask questions when you already know the answers and they're irrelevant.
>
> When you say f(x) = something, it is not the same as
> y = something unless you also define y = f(x). That is
> standard practice beginning, I'd say, by about 11th
> grade.
What "y" were you asking about? I didn't bring up 'y'.
>
> And when you say f takes an argument you call n, then
> no it is not assumed that x stands for the argument
> which just before was called "n". That's a pointless
> confusion.
>
f(n)=f(x)=f(codpiece) for n=x=codpiece. Why so easliy confused?

> > > Is f(n) taken by axiom to be a bijection? How do you know
> > > it always exists?
> > Bijections are irrelevant to what I am doing.
>
> You are assuming that f(n) is invertible. That implies
> certain properties. Not all functions are invertible.
> Here's one that isn't: f(n) = abs(n-6).
I stated that restriction. What does Cantor have to say about such sets?
They're all the same. Verrry interesting result.
>
> > To answer your question in real words, I think you are asking whether
> the
> > output of the function is always a natural.
>
> Then what's wrong with x={1.3,2.6,5.7}?
>
> > > Well, I can't call it N, since N is a set and not a number.
> > It is both.
>
> No, it isn't.
>
> > We're not in Cardinality Land anymore, Randy.
>
> We're trying to be in logic land, where symbols have fixed
> meaning.
>
> > Does |N| make it clearer?
>
> Yes. Editorial comments deleted.
>
> > Who needs a new special symbol, except to make things more
> mysterious?
>
> Introducing |N| to talk about the size of N is neither
> mysterious, new, nor special. It has intuitive connection
> to the magnitude |z| of an arbitrary (real or complex)
> number.
Fine. Context is everything, but I can give you bars for clarity.
>
> > > Is aleph_0 - 1 < aleph_0?
> > Yes, correct. A proper subset is always smaller than the superset.
>
> The big question is whether a set has only one cardinality
> in your scheme. You know that more than one f(n) can
> exist, right?
for a particular n and f? I think each set has a single cardinality, though
this set of finite naturals is problematic to think about, and indefinite in my
mind. It tastes bad in my mouth.
>
> > > No, I'm convinced that a plausible order of "bigness"
> > > can be constructed this way to compare some subsets of the
> > > naturals to the set of naturals. I'm not convinced
> > > you can compare all sets this way, nor am I convinced
> > > that you can even compare all subsets of the naturals
> > > to each other.
> > I can order all subsets of the reals that are defined using an
> invertible
> > function on the naturals in this way. You will notice that using this
> method,
> > there is no function, like 2^N, as Cantor suggested, that states the
> > relationship between the naturals and reals.
>
> Well it arises naturally when considering the reals in
> [0,1].
Well it arises naturally when enumerating the naturals too, if you use base 2.
If you use base 10, then there are 10^N naturals. There IS another way to view
the reals, where they seem to have 2^N members regardless of any chosen number
base, but that still doesn't satisfy me as an answer to that question.
>
> How many bit sequences of length 5 are there? 2^5
>
> How many bit sequences of length n are there? 2^n
>
> How many bit sequences of length |N| are there?
2^N
How many decimal sequences of length n are there?
>
> Since you believe that Mueckenheim's counting method for
> the binary sequences generates all of them, even the
> infinite ones, surely that set has cardinality 2^|N|
> by your count?
2^N reals? I think he was talking about reals between 0 and 1, so it might be N
times that overall. Still, I don't believe that constitutes a careful enough
definition of the size. Using my mthod, any such finite function will result in
a less than full population of the real values, so I don't believe any finite
function can define the ratio between |R| and |N|.
>
> > That is, there is no finitely long
> > function on the naturals which completely fills the real line with
> values.
>
> What's a "finitely long function"?
A function defined in a finite number of symbols. The only possiblity I can see
for a function to define all R is something like:
log(log(log(log(.....x))))....)). This would progress infinitely slowly and
cover all the reals, but is what I would call an infinitely long function. Does
that make sense?

>
> > > Is the set of odd numbers bigger or smaller than the
> > > set of even numbers?
> > The same. Half of all whole numbers. You see (crank alert, yeah yeah,
> I know)
> > infinity is odd, which balances zero's even. There is a reason out
> there,
> > somewhere waiting to be found, that infinity is prime, but it is
> still eluding
> > me. I hope that gave you a big laugh.
>
> No laughs. I find ignorance-by-choice to be very sad,
> actually.

So do I. It's the highest spiritual crime.

>
> > I hope more that it actually made sense
> > to you, but I doubt it.
>
> Oh it made sense. I recognize the naive errors you're making,
> and I know the intuition you're trying to appeal to. I
> also know that it leads to no useful, self-consistent
> notion of set cardinality. So it's not that it doesn't
> "make sense to me", I know that it's not a self-consistent
> theory, period.
Please state the contradictions. If it doesn't contradict anything you hold
dear, you have no complaints, except maybe you don't see the need for it. So,
don't use it.
>
> There is a well-grounded, mathematically rigorous
> concept called "natural density" which is definable for
> many (but not all) subsets of the natural numbers.
Yes, I know. I started with that and generalized it to the method we are
discussing. It always deals with the entire infinite set and measures it over
that range to N. Finite fractions of N are given finite numbers. Infinite sets
that are infinitesimal relative to N, such as the set of squares, are given the
value zero. Not satisfactory to me. I improved it. Thanks for the tip.
>
> > > How big is the set of rationals on [0,1]? Here's an
> > > f(n): f(n) = 1 - 1/n. Using that, what do you get
> > > for your "bigness" of the set of numbers {1-1/n}?
> > > How about the "bigness" of the set of all rationals
> > > 1-1/n?
> > This method doesn't address the rationals, or any other dense
> ordering of the
> > reals. Dense orderings cannot be defined by finite functions on the
> naturals.
>
> But I just defined a function to generate a particular
> subset of rationals. Doesn't that subset have a
> Tony-ality?
yes. You snipped that part. N-1.
>
> This is way too long now. Will read and respond to the
> rest in a bit.
Oh you didn't get to it yet. Hmmm read on.....
>
> - Randy
>
>

--
Smiles,

Tony
.

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