Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
cbrown@xxxxxxxxxxxxxxxxx schrieb:


This bijection between sets (initial segments {1,2,3,...n} and
{2,4,6,...2n}) is only valid for finite sets.

Suppose I instead counter-claim that this bijection is obviously "not
valid" for finite sets; only for infinite sets.

Is that what you call a "correct mathematical argument"? Simply
claiming something is valid or not valid? How is it different from your
own argument?

It would be as impossible to contradict as your other claims...

So you agree that that /is/ what you call a "correct mathematical
argument" - simply claiming something is valid or not valid.

Of course it will be difficult to refute statements made with this
justification. In the words of South Park's Towelie: "No man, /you're/
a towel!"

or as the
following:
In every finite initial segment of natural numbers, there are as many
even numbers as odd numbers (plus or minus one number). But in the
infinite set N there are 250 times more even numbers than odd numbers.

If you specify what you mean by "250 times more", then your statement
will have meaning; until then, it is nonsense.



Thus "|N|/|R| < 1" is nonsense; as I previously stated
quite clearly.

I did not calculate |N|/|R| but edges per path.

Wonderful. But what relevance does the calculation of "edges per path"
have to do with the /actual question/: does there exist a surjective
function from the set N to the set R?

That is what is meant by "|N| >= |R|". The use of ">=" here has a
/different meaning/ than "1 + 1/2 + ... >= 1", a difference which seems
to elude you.

In the latter case, we are not asking "is there a surjective function
from 1 + 1/2 + ... onto 1?", because that would be simply nonsense.
Similarly, to try to show that |N| >= |R| by calculating the limit of
edges per path is equally nonsense - it requires misreading the meaning
of ">=".

This calculation yields
a real number for any finite path - and the infinite path is nothing
else but the union of all finite paths.

And therefore... the union of these calculations is a countable set of
real numbers? And that shows a surjection from N to R how?


The number of edges accumulated by one path up to level n is a function
like that above:
f(n) = 2 - 1/2^n. Very easy.

Yes; when you write it that way, it "looks" just like a valid statement
about the real numbers. So, you "feel" that it is a both a sensible and
a valid statement about things which are not real numbers; when instead
it is nonsense.

What is your definition of an infinite path?

There is a countable set of edges. For each natural number n, there is
a distinct edge which is the nth edge in the path p. If an edge e is
the nth edge in the path p, then there is no other natural number m
such that e is also the mth edge in the path p. An edge e is part of a
path p iff there exists a natural number n such that e is the nth edge
of p.

More concisely: |edges| = |N|. A path p : N -> edges is an injection.
An edge e is part of path p iff e is in the image of p.

You feel that it is not
the union of all finite paths...

Where did I state that? Trivially, each edge in a path is a finite set
with one element (the edge). The union of all singleton sets of edges
is of course the same set as the set of all edges (i.e., the infinite
path). The set of all singletons edges is a subset of the set of all
finite paths; thus the union of all finite subsets of a path is the
infinite path.

Thus it's not a matter of what I "feel"; I just /proved/ that the
infinite path is the union of all finite subsets of that path.

But what is your point? An infinite path is also the union of all
infinite proper subsets of that path. So what?

but something much much greater and
longer, more exciting and most mysterious?

There really is nothing at all mysterious about it. It simply follows
from the definitions of N and "an infinite path". What is mysterious is
why you continue to claim to find a surjection from N to R.

Do you have a definition how
or any reason why the infinite path is not the union of all finite
paths?


No. Have you given up on your "rational relation" proof that |N| = |R|?

Cheers - Chas

.

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