Re: Cantor Confusion

Virgil wrote:
In article <1164446864.918062.126880@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

In article <1164307852.961046.140060@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

I will accept that it can be connected if the "right" uncountable
set is
removed, but I can think of uncountable sets of points whose
removal at
least appears to totally disconnect the set of those remaining.

Of course, for example if you remove all points.

The remaining empty set is trivially connected.

That is nonsense. If there is nothing to be connected then there is
nothing connected!

Topologically, a set is connected unless it is a union of two or more
non-empty disjoint open sets.

See http://en.wikipedia.org/wiki/Connected_space

According to this definition, an empty set is connected.

I prefer to think myself.

If WM prefers to substitute his own definitions in place of standard
mathematical definitions, his version of "mathematics" will necessarily
and obviously be irrelevant to any standard mathematics.

The point of having a standard mathematics is that everyone must agree
to use the standard definitions in order to discuss it reasonably.

Those who insist on using non-standard definitions are required to give
mathematically adequate definitions of what THEY mean if they expect
their definitions to be understood or acceptable.

What definition of connectedness does WM use which disconnects the empty
set?

A set is connected, if it has at least two elements and if its elements
are all connected.

What constitutes a connection versus a disconnection? Topology has a
clear definition. You do not. Ergo, your "definition" is irrelevant.


A set with only one element cannot be connected or
disconnected.

Why? Topology has a precise answer, Do you?


A set with less than one element is, strictly speaking,
not a set at all.

It is in ZF and NBG. Does WM then wish to claim that the intersection
of two sets is sometimes "not a set at all"?


So it is neither connected nor
disconnected.

WM, on the other hand, is totally disconnected on such issues.




With other words, you did not understand them. But our question
concerned the appendix only. The proof given there is very simple, so
simple that even Mr. Bader asserted to have understood it (it seemed so
to him, at least). You cannot confirm that my proof is correct?

In reading the beginning of that paper, I found enough errors to
dissuade me from bothering.

So it should be possible to name one, unless you prefer to join the
slanderer, Mr. Bader.

It is asserted in that paper that the so called first Cantor proof of
the uncountability of the reals applies equally well to the set of
rationals.

Cantor's proof relies on the fact that if one has a strictly increasing
sequence of reals with each term less than all terms of a strictly
decreasing sequence of reals, there is at least one real strictly
between the two sequences, not being a member of either but larger than
every member of the increasing sequence and smaller that every member of
the decreasing sequence.

This is not the case for rationals. For example, there are strictly
increasing sequences of rationals whose squares converge to 2 and
strictly decreasing sequences of rationals whose squares converge to 2.
But there is no rational number between the sequences.

For example a_1 = a, a_{n+1} = 4*a_n/(2 + a_n ^ 2) is an increasing
sequence of rationals and b_1 = 2, b_{n+1} = (2 + b_n ^ 2) / (2 * b_n)
is a decreasing sequence of rationals with only sqrt(2) between all the
a_n and all the b_n.


Regards, WM

PS: You really did not see that Cantor's arguing in his first proof is
valid for rational numbers as well as for transfinite numbers?

Not so, as the example above proves. There is an increasing sequence of
rationals converging to sqrt(2) and a decreasing sequence of rationals
converging to sqrt(2) such that there is no rational at all caught
between the two sequences.

If WM thinks this, he has not understood the first proof.

There is an increasing sequence of rationals whose LUB is sqrt(2),
e.g., f(n) = floor( 2^n * sqrt(2) ) / 2^n.

And a decreasing sequence of rationals whose GLB is sqrt(2),
e.g., g(n) = Ceiling( 2^n * sqrt(2) ) / 2^n

If Cantor's first proof held for the rationals, as WM claims, then WM
would have proved that sqrt(2) is rational.

You misunderstood completely.

No, it is you who misunderstand Cantor's proof.

I claim that Cantor's proof is
"symmetric". It can be applied to the algebraic numbers, showing that
the limit is not algebraic.

It can be but need not be.

It can be applied to any proper subset of the reals showing that there
is a sequence of values in that subset which does not converge to a
value in that subset. So that anything short of the full set of reals is
not closed under Cauchy convergence. One definition of the real number
system is the closure of the rationals under Cauchy convergence.



It can as well be applied to the
transfinite numbers, showing that the limit is not transfinite.

This I would like to see. A sequence of non-finite numbers having a
finite limit? By what definition of limit does this sequence even have
a limit? For sequnces of real numbers one can use a delta-epsilon
definition or a topological definition of limit, but what



This
has nothing to do with rational numbers and sqrt(2) because that would
not prove any uncountability at all (sqrt(2) is algebraic and as such
belongs to a countable set).

The point is that for every sequence of reals, there are reals not in
that sequence.

For a set to be countable, it is necessary and sufficient that there be
a sequence containing every member of that set.

This can be done, and has been done, with the set of rationals and with
the set of algebraics, but not with the set of reals. It CANNOT be done
for the reals.

Cantor's first proof shows that ANY sequence of reals omits at least one
real, thus any attempt to ennumerate the reals, in the sense of
surjecting the naturals onto them, fails.

That's not so where the reals, non-standardly or whatever, can be
addressed as contiguous points on a line.

Megill's proof checker found the rationals uncountable.

Well-order the reals, biject them to an ordinal. Does Cantor's first
in transfer apply? At some point in time there's a degenerate
interval, and before that not, extend inductive reasoning to infinite
ordinals, oops, now you're talking about the set of naturals as if it
were an element, of itself, for inductive purposes.

There is no universe in ZF. There isn't a set of sets in ZF, there are
no sets in ZF, identity is too big.

Cauchy/Dedekind definitions (acknowledgment, placement, in context) of
real numbers are: not sufficient.

Biject R[0,1]^x to N^N and sample towards an N-sequence, that appears
to be a random natural integer.

It is you, Virgii, who misunderstand, and so vociferously, and much.
That's not to say others don't: you do.

Ross

.

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