Re: abundance of irrationals!)


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.

> 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"?

All I gave you was the computer algorithm for finding
the max of an unsorted list. It doesn't require you
to know the largest element in advance.

> > 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....

Well, yes. That set doesn't have a largest element either.

> > 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?

I want to reserve "finite" for sets whose members can be
numbered from 1 to a natural number n.

> 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.

Will counting the elements ever stop?

If it stops, call it "finite".
If it doesn't stop, call it "infinite".
Is there a third possibility?

> > 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?

That the Tony-ality of a set may depend on the ordering
of the mapping function.

> > > 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.

Well then, where did I say that finite sets by definition
have largest elements?

> > 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.

I have to define an ordering function that lets me
compare any two cards.

> You have a 2D matrix, 4x13=52. Rationals have a 2D matrix as well:
> NxN=N^2 rationals.

But there exists a way of lining them 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.

> > 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".

> > 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.

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 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.

The thing in quotes was taken by cut and paste from your
post. So the person who left that out was you. Hence
my statement that you left it 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.

Nothing magic about 1,2,3 as the set I am mapping from.
Nor about the order. I could write down a different
set and consider the cardinality of the set {f(0),
f(1000), f(6)}.

The inverse function will give me the three values
0, 1000, and 6. How does your cardinality function
work then?
> > 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.

I quoted you what you wrote up front. You didn't like it,
informed me it was missing stuff. QED.

> 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?

Yes, I said that before. It's related to "natural density",
but as a cardinality measure it has a lot of problems.

> > 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?

I'm not yet clear that it does so in a consistent manner.

> 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'.

Should I quote again what you wrote as your definition? It
starts out this way:
"The size of a set of real numbers defined using a function f(n) on
the naturals from x to y..."

That's the y I was asking about.

> f(n)=f(x)=f(codpiece) for n=x=codpiece. Why so easliy confused?

It's pointlessly confusing to tell me f(n) is from
x to y. When we talk about what a function is "from"
and "to" we mean the domain and range of that function,
not the arguments. So when you said "f(n) is from x to
y" I thought x and y were sets, in the standard language.

> > > > 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.

What sets?

> > 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.

Which set?

Here's a set: {1.3,2.6,5.7}. I can define f1 that
generates this from {1,2,3}. I can define f2 that
generates this from {1000,1,67}. Both are invertible.

I can define f3 that generates it from {3,2,1}. So yes
there is more than one function f(n) that can take you
from {1,2,3} to {1.3,2.6,5.7}, by mapping them in
different orders.

- Randy
> > 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?

10^n

> > No laughs. I find ignorance-by-choice to be very sad,
> > actually.
>
> So do I. It's the highest spiritual crime.

Irony meter pinned.

- Randy

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