Re: how

On 4/17/24 2:52 PM, WM wrote:

Le 17/04/2024 à 01:29, Richard Damon a écrit :

On 4/16/24 2:36 PM, WM wrote:

Le 16/04/2024 à 18:38, Mike Terry a écrit :

On 16/04/2024 16:05, WM wrote:

Le 16/04/2024 à 02:35, Mike Terry a écrit :

On 15/04/2024 14:00, Tom Bola wrote:

>

0, 1, 2, 3, ...,   w,   w+1,   w+2,   w+3, ...     w+w
|  |  |  |  |||    |     |      | |            |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2

>
Should be:
>
0, 1, 2, 3, ...,   w,   w+1,   w+2,   w+3, ...     w+w
|  |  |  |  |||    |     |      | |            |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2

>
No. (ω+1)*2 = ω*2 + 2

>
No, you need to learn how ordinal arithmetic works:
>

I see: 1 + ω = ω.
>
Nevertheless the question remains where in the second row is ω located, doesn't it?.
>

Why does it need to be there?

 Because ω is smaller than ω2.

So, the bottom row is missing all the odd natural numbers. why can't it skip all the odd multiples of ω too?

>
THe two set/series are of the same size even though the bottom misses the "odd" values.

 The two sets are of same number of elements because every elements is doubled. But after multiplication these elements cover an interval twice as large as before. The end is ω2. Therefore ω or its neighbours are located in the midst of the interval.

Nope, unless by "midst" you mean that one is above it and the rest below it.
Note, your original set wasn't "equally spaced" as the ω at the end is an infinite distance from the rest of the set, so you can't use equal spacing logic on the second set.

 Regards, WM