Re: Two results of set geometry

On 11 Sep., 14:00, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Sep 11, 7:46 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:





On 11 Sep., 13:33, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

You have the full column. It is the union of the finite segments.

And this union is infinite.

Correct. You now see that it is possible to have an infiniteset
made up of finite things.

I never denied that. But it is impossible to have an infinitesetmade
up by different finite things which differ by a fixed amount like 1.

Please don't mistake the one for the other.

Note that each finite segment differs by the fixed amount 1.

So it is. And what can we say about the (completed) infinite set?
It contains an infinite segment! At no other point infinity is completed.

The two statements

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

are not contradictory.

No? There is no contradiction if we say:
The set A is complete and the set A is not complete? (What else would
it mean to say there is no point at which the set A is completed. Why
should A have a point where it is complete?)

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, ...

The set of all finite initial segments
is complete (it contains all finite initial segments). However, there
is no segment at which the set is completed.

The infinite segment is, by definition, the union of the finite
segments.

And which one is omega?

An infinite set made up of finite things which differ by a fixed
amount like 1 is impossible. Think simply of a staircase: An infinite
set of steps implies an infinite height.

Or do you insist that an infinite number of steps of height 1 does not
imply an infinite height?

Regards, WM

PS: I cannot see your previous contribution, only the announcement.

.

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