Re: Cantor Confusion

> So the union of finite trees U(T(n)) contains (as subsets) the path
> p(oo) and all its co-paths q(oo), ..., i.e.,

Yes. I never said otherwise. Why do you think I said otherwise?
All the p(oo) are subsets of U(T(n)).

> it contains P(oo)?

And that is wrong. Pray look close at what the elements of the different
sets are:
U(T(n)) has as elements nodes

and it has paths as subsets.

P(oo) has as elements paths, i.e. sets of nodes
so P(oo) is neither an element of U(T(n)), nor is it s subset of it.

Every element of P(oo) is a path and as such a subset of U(T(n)), as
you say.
U(T(N)) has only finite paths as subsets (as N has only finite initial
segments --- both U(T(n)) and N have infinitely many such subsets,
but these subsets are finite).

Eh? Above you state (properly) that p(oo) is a subset of U(T(N)). Are
you contradicting that now? And if we consider: N subset N, N contains
one infinite initial subset. Or do you mean something different with
U(T(N)) from U(T(n))? If so, what do you mean with it?

There are countably many unions of finite subsets of U(T(n)).

Prove it. There are countably many *finite* unions of finite subsets,
not countably many arbitrary unions of finite subsets. Consider N.
Each subset of N is a union of finite subsets of N, namely the union
of all singleton sets that contain an element of that set.

There are countably many unions like U(p(n)).

Prove it.

Therefore there are countably many elements p(oo) = U(p(n)) of P(oo).
Therefore, P(oo) is a countable set.

Wrong.
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