Re: Non-denumerable ordinals

Claim 2. There is no injection from W to S.

I am sorry I did not pay much attention to your proof, David, since I
wished to find one myself.

Of course, the natural approach is to try to find a contradiction
from:

There is an injection i from W to S.

Range(i) is a subset of S well-ordered by the relation

R = {<x,y> of Range(i) X Range(i) | i^(-1)(x) < i^(-1)(y)}

where < is the well-ordering relation

< = {<w,w’> of W X W | Some u of w is isomorphic to an initial segment
of some v of w’}.

<Range(i),R> is, then, an element of some equivalence class w of W.
And that w has an image i(w) of Range(i).

The same argument works for every subset of Range(i) since it is well-
ordered by the same relation R. Particularly, it works for every
initial segment of <Range(i),R>.

For each a of Range(i) we have an initial segment Range(i)[a] = {x of
Range(i) | x R a}. Therefore, we have a one-to-one function j from
Range(i) into W.

Range(j) is a subset of W, well ordered by <. What’s more, j is an
isomorphism between <Range(i),R> and <Range(j), < >.

Since i is an isomorphism between <W, < > and <Range(i),R>, by
composing i and j we get an isomorphism between <W, < > and a subset
of <W, < > this is impossible unless the isomorphism is identity, i.
e. unless j is i^(-1).

Now, what is the image under i of the equivalence class of <Range
(i),R>? We know now that for each element a of Range(i), its pre-image
is the equivalence class of the initial segment Range(i)[a]. Since for
no a can Range(i)[a] be isomorphic to Range(i), the latter cannot be a
member of the corresponding equivalence class. Therefore, there is no
image under i of <Range(i),R>. A contradiction.

.

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