Just below first inaccessible cardinal

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tzimie
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First inaccessible is "above" all [tex]\aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}[/tex] etc.

My question is what is the "height" of that tower, can it be not only infinite, but also as huge as [tex]\aleph_{\aleph_{\aleph_...}} \big\} \aleph_{\aleph_0}[/tex] so the height of the tower is defined by another huge cardinal?

Does adding more levels to that structure (cardinal defines the height of a tower of another cardinal, which defines the height a tower on level 2, etc)

[tex]\aleph_{\aleph_{\aleph_...}} \big\} \aleph_{\aleph_{\aleph_...}} \big\} \aleph_{\aleph_0}[/tex]

bring anything new, or any "multilevel" cardinal can be defined as using a single level notation?
 
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tzimie said:
First inaccessible is "above" all [tex]\aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}[/tex] etc.

My question is what is the "height" of that tower, can it be not only infinite, but also as huge as [tex]\aleph_{\aleph_{\aleph_...}} \big\} \aleph_{\aleph_0}[/tex] so the height of the tower is defined by another huge cardinal?
I don't know about the notation using the symbol N0, N1, N2 etc. This cardinality topic is somewhat beyond my knowledge. Probably someone more knowledgeable and/or comfortable with this topic can give a better response (hopefully my response doesn't contain some major inaccuracy). But nevertheless, still here is my thought on it.

I think it should make sense. Simply thinking about it in terms of Ω (first uncountable). Let's write Ω as Ω0. Let's simply define a function f (from ordinals to ordinals I guess?) such as:
f(x)=Ωx=x-th uncountable ordinal

Now assuming that the fixed points for this function work in the same way as for countable ordinals (f should be a normal function), we have its first fixed point as the supremum of (fn denotes composition of f n times):
A={f(0),f2(0),f3(0),f4(0),...}

And this essentially corresponds the "height" of the tower you described being equal to ω. If we then denote g as the function enumerating fixed points of f, then we can also write something like:
g(ΩΩ0)=g(f2(0))
it essentially corresponds to the term in quote above. Only sort of ... I think(?) because N0 is equal to ω and N1 is equal to Ω?. In that case we can just write: g(Ωω).
 
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