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Would uncountable infinite^uncountable infinite^uncountable infinite^uncountable infinite (repeat uncountable infinite amount of times) be enough to reach Aleph 2?
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Question about Aleph numbers
- Thread starter Lord_Zeref_Dragneel
- Start date
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Depends on context.
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2^countable infinity is Aleph one. no idea about this one tho
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What. No it is not. You cannot reach Aleph 1 by simply stacking infinities.
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To whom are you referring to?
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according to the continuum hypothesis, it just works.
Cardinal
View full site to see MathJax equation In set theory, the cardinal numbers (or just cardinals) are equivalence classes defined by the relation "there exists a bijection from set \(A\) onto set \(B\)".[1] Whereas ordinal numbers may be thought of as "structures" of certain kinds of sets...
refer to the continuum hypothesis section.
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stacking finite numbers won't give u infinite numbers
stacking countable infinite numbers won't give u uncountable infinite numbers
and so on, stacking aleph one won't give u aleph two
but the power set of a cardinal will always be bigger than the cardinal itself
so 2^aleph null = aleph one
2^aleph one = aleph two
and so on
in your case, yes, this is aleph two.
stacking countable infinite numbers won't give u uncountable infinite numbers
and so on, stacking aleph one won't give u aleph two
but the power set of a cardinal will always be bigger than the cardinal itself
so 2^aleph null = aleph one
2^aleph one = aleph two
and so on
in your case, yes, this is aleph two.
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Idk I always heard that simply stacking infinities even if in power sets won't take you to the next level
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power sets can
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I see. Guess all the info I had were wrong then lol. Thanks
Cat275
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Power sets and collection of subsets is always bigger than a cardinal.
I know this is kinda late
But in the continuum hypothesis assuming it's true it would work like this
2^Aleph null=Aleph one
And then just continue endlessly.
Anyways about the question it should be aleph two, yes.
Note:You can't reach limit ordinals or cardinals by this kind of sequence you would need a different method or atleast different in a way for some.
(Mostly large cardinals honestly.)
I know this is kinda late
But in the continuum hypothesis assuming it's true it would work like this
2^Aleph null=Aleph one
And then just continue endlessly.
Anyways about the question it should be aleph two, yes.
Note:You can't reach limit ordinals or cardinals by this kind of sequence you would need a different method or atleast different in a way for some.
(Mostly large cardinals honestly.)
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