Re: infinity

Nathan said:
> Tony Orlow (aeo6) wrote:
>
> > David R Tribble said:
>
> > > It also demonstrates that the two sets have exactly the same number
> > > of members; any given member of one set will have a numeric value
> > > that is exactly twice or half of the corresponding member in the
> > > other set.
>
> > That's not at all the way I interpret it. If the two sets have the same range
> > of values, then the one with the lower density or frequency will have fewer
> > members. When you draw a bijection between the two sets using a mapping
> > function that doubles one to get the other, you are also doubling the value
> > range of the first to get the value range of the second, so you get the same
> > number of elements in twice the space. If the two sets cover the same range,
> > then they will not have the same number of elements.
>
> What about sets that don't contain numbers? An important part of the
> concept of the cardinality of a set is that it doesn't matter what
> characteristics the set's elements have. {1,2,3}, {x,y,pi}, and
> {Alice, Bob, Carol} are sets with the same cardinality. You can change
> the "names" or characteristics of the elements, and the cardinality
> stays the same. You want this to work differently for infinite sets?
> What, nothing but numbers can be elements of an infinite set?
When working with infinite sets, there is really no way to compare the sizes of
sets without resorting to properties of their elements. Numeric sets are one of
the most common, but discussions of infinite sets also involve structures such
as trees, processors such as Turing machines, and infinite systems such as
symbolic languages, which include symbolic number systems. When drawing a
bijection using an arithmetic formula, we are dealing with sets of quantities,
and inverse functions are the way to compare those kinds of sets. When working
with symbolic systems of any sort, the formula N=S^L is used, where N is the
size of the maximal language given a symbol set of size S and string length of
L. In general, this yields a maximum size of the language. For digital systems,
every string is included and has a unique value, so this formula becomes a
quite exact way to gauge those infinite sets. For binary trees, we need to
closely examine the structure of the tree at the node/branch level to gauge the
relationship between infinite nodes, branches and paths, as we discussed at
length in Meuckenheim's Cantor and the Binary Tree thread (where he was more or
less correct, by the way, despite being accused of quantifier dyslexia and
general idiocy). So, in my opinion, the desire to have one simple method for
comparing all infinite sets is unreasonable, since there ARE different kinds of
infinite sets with different element properties that need to be examined.
>
> When I visualize a line segment, say the interval [0,1], I may draw a
> short line on a piece of paper to represent what I'm thinking. I may
> even label the endpoints "0" and "1" to be clear. If I want to change
> my diagram to represent [0,2], I can erase 1 and write 2. I haven't
> changed the size of my diagram, and I haven't changed the cardinality
> of the set--I've just "relabelled" them.
You have then halved the unit of measurement you are using. If you called it
inches, in at least one case your diagram would be wrong.
>
> In using the function x -> 2x to create [0,2] as the image of [0,1],
> how can the "number" of elements have increased? Each element of
> the domain produces exactly 1 element of the image, and each element
> of the image comes from exactly 1 element of the domain. There's no
> way for any "extra" elements to "sneak in" to the image from anywhere.
> >From a certain point of view the elements have just been "relabelled".
> It's still the same line I drew on my paper.
When you do that, you have an interval [0,2] that is half the density of the
interval in [0,1]. You are basically creating a bijection between
{0.000...000,0.000...001, 0.000...002, etc} and {0.000...000, 0.000...002,
0.000...004, etc}. If you specify that both intervals are of the same density
(say, of the continuum), then the size of the set in terms of points is
measured by the length of the interval. After all, [0,2] contains ALL points in
[0,1], plus the infinity of points in (1,2]. How can this additional infinity,
equal to the first (minus 1 point), NOT increase the size of the set? The set
has doubled.
>
>

--
Smiles,

Tony
.

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