Re: Two results of set geometry

On 12 Sep., 02:10, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

Yes. However, the fact that an *infinite set* of steps
has infinite height, does not imply that any single step
has an infinite height.

What is it then that causes infinite height?

If a step is not the last step, the set of steps has a height greater
than that step.

The set of steps has no height at all. It has a number of elements.

No step is the last step, so the set of steps has greater
height than any step.

The set of steps has no height at all. A set of green elements is not
green! A set of large elements need not be large.

For any natural number there is a step of that
height.

The set of steps has height greater than any natural number.

The set of natural numbers is not a natural number. The set N has
omega elements but it has no numerical seize.

The set of steps has infinite height.

It has no height.

There is no last step.

But all steps which are there, are finite. Therefore they do not cause
infinite height.

This stands unrefuted.

Back to the subject.

It is not contradictory to say

The set A is complete.
There is no point at which the set A is completed.

It is a contradiction. In order to say that A is complete, you must
know all elements of A.

No. To say that A is a complete set of things of type X, you
have to know that given a thing of type X you can show that it
is an element of A.

That is a tautology. The statement "given a thing of type X" implies
that that the thing belongs to A. There is nothing to be shown. But
even what you mean is not enough. You have to know that you are able
to consider all elements of the set A simultaneously (simultaneous
being together of elements is essential for sets and is impossible for
some non-sets, as already Cantor knew). This is only possible if you
know that there is no element left out.

You don't have to know all things of type X,
it is enough to know something that must be true of any thing of type
X.
Knowing all the elements of A is neither necessary nor sufficient.

Why then is the set of all sets a problem?

It is not contradictory to say

The set A is complete.
There is no point at which the set A is completed.

It is contradictory. But in order to reduce the argument to a rational
level take my recent argument:

Does the sequence S = 111... represent omega? Does it contain an
infinite segment?
Or do the of finite segments of S, namely 1, 11, 111, ..., represent
omega?
Or does counting stumble somewhat when "counting over into the
infinite"?
1, 2, 3, ..., omega, omega, omega + 1, ...


Regards, WM


.

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