High School Subsets of Rational Numbers and Well-Ordered Sets

[SSequence]

This isn't original or anything, but I was thinking about how would one go about formalizing (in a general sense) an informal wikipedia picture such as this:
https://upload.wikimedia.org/wikipe...led.svg/487px-Omega-exp-omega-labeled.svg.png

For example, some people might have seen similar pictures for ω². So it seems interestingly there is a fairly reasonable correspondence between the kind of pictures posted above and what I describe below:
Place the number 1/2 at 0
Place the number 1/4 at 1
Place the number 1/8 at 2
and so on...
We can also add the assumption that the number placed on each position must be strictly positive.

But we want to distinguish between "number placed" and "number assigned".Now we can define the "number assigned" for each position as:
The running sum of numbers placed at all lower positions.

So the (unique) "numbers assigned" in the above case are:
0 is assigned 0
1 is assigned 1/2
2 is assigned 1/2+1/4
3 is assigned 1/2+1/4+1/8
... and so on.
1 is assigned to ω

But we can go beyond easily here of course. For example, going a bit further to see this:
Place the number 1/4 at ω
Place the number 1/8 at ω+1
Place the number 1/16 at ω+2
Place the number 1/32 at ω+3
...

So now we have unique numbers assigned as:
1 is assigned to ω
1+1/4 is assigned to ω+1
1+1/4+1/8 is assigned to ω+2
1+1/4+1/8+1/16 is assigned to ω+3
... and so on
1.5 is assigned to ω.2

Now indeed we can generalize this suitably to assign unique rational numbers to all elements less than ω². For example, following the pattern, we would have 1.75 assigned to ω.3 and 1.875 to ω.4.

The main point here is that the "numbers placed" must be such that the running sum doesn't go unbounded. Now what is interesting in this kind of assignment is that if r1 is the unique rational number assigned to ordinal t1 and r2 is the unique rational number assigned to t2, we have:
(r1<r2) only if (t1<t2)

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One point here seems to be that we must have running sum as convergent for each limit. And furthermore, it seems that if the convergence value is not a rational number on some limit, we might have some further worries to attend to. I don't know about the generic conditions when the convergence value of a convergent rational sequence is itself supposed to be rational.

 

[Mark44]

One point here seems to be that we must have running sum as convergent for each limit.

I don't see that as a problem, since in each of your examples you are adding terms of a convergent geometric series.
To get 1.5, you have 1 + 1/4 + 1/8 + 1/16 + ... + 1/2ⁿ. The terms starting with 1/4 converge to .5
To get 1.75, you apparently have 1.5 + 1/8 + 1/16 + 1/32 + ... + 1/2ⁿ. The terms starting with 1/8 converge to .25.

 

[SSequence]

Yes it seems that in specific cases one can always choose the values (for "numbers placed") such that the unique assigned numbers converge (and to a rational number) on each limit. In fact, it seems that after suitably formalising this one can show (using induction for example) that this can be done for every countable (one would probably have to be a little careful about upper-bounds here).There also seems to be one other aspect. To be more specific, considering the example that I posted in first post. Consider the unique assigned numbers corresponding to the following ω-sequence:
0, ω, ω.2, ω.3, ...
We have the corresponding assigned numbers as:
0, 1.5, 1.75, 1.875, ...
This sequence of values converges at 2, which would be the same as the assigned number to
ω² (defined originally in terms of running sums).

So it seems that one would want to consider the following two results (perhaps both combined as one) to make things easier:
(i) The unique number assigned on each limit value (say t) is same as one obtained by finding the limit of the sequence of assigned numbers (corresponding to some ω-sequence of p).
(ii) Given any two different ω-sequences for p, if we consider the sequences of corresponding assigned numbers, then if anyone these sequences converge then both of them do (and to the same value).