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Is Aleph one should be considered to be uncountable Infinite rather than Infinite^Infinite?
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Is infinite^Infinite considered to be aleph null? And it's not uncountable infinite
- Thread starter RamadhanDayYet
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Hasty12345
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from my understanding, Aleph_1 = 2^(aleph_0) via the continuum hypothesis,
2^(aleph_null) <= (aleph_null)^(aleph_null)<=(2^alepth_null)^(alepth_null)<=2^(2*aleph_null)=alepth_1
the first step, 2^(aleph_null) <= (aleph_null)^(aleph_null) just comes from infinity being larger than 2, and since we're assuming it's greater than or equal to 2^(aleph_0), we can sub it into the next step to prove they are equal.
Then "infinite^infinite should be alepth_1
There could be a mistake, I always hated analysis
Edit: after doing some research, the continuum hypothesis is currently "undecided", meaning 2^aleph_null = c = cardinality of R which may or may not be aleph_1, apparently the CH cannot be proven or dis-proven within the standard axioms of mathematics.
I found this wikipedia article to be helpful along with these:
en.wikipedia.org
www.ias.edu
the biggest issue of the CH from my understanding is proving if there is a cardinal between aleph_null and aleph_1, so aleph_null<cardinality of R which may or may not be = alepth_1.
I'm sure there are many people far more knowledgeable than me who can help you.
2^(aleph_null) <= (aleph_null)^(aleph_null)<=(2^alepth_null)^(alepth_null)<=2^(2*aleph_null)=alepth_1
the first step, 2^(aleph_null) <= (aleph_null)^(aleph_null) just comes from infinity being larger than 2, and since we're assuming it's greater than or equal to 2^(aleph_0), we can sub it into the next step to prove they are equal.
Then "infinite^infinite should be alepth_1
There could be a mistake, I always hated analysis
Edit: after doing some research, the continuum hypothesis is currently "undecided", meaning 2^aleph_null = c = cardinality of R which may or may not be aleph_1, apparently the CH cannot be proven or dis-proven within the standard axioms of mathematics.
I found this wikipedia article to be helpful along with these:
Cardinality of the continuum - Wikipedia
Can the Continuum Hypothesis Be Solved?
In 1900, David Hilbert published a list of twenty-three open questions in mathematics, ten of which he presented at the International Congress of Mathematics in Paris that year. Hilbert had a good nose for asking mathematical questions as the ones on his list went on to lead very interesting...
www.ias.edu
the biggest issue of the CH from my understanding is proving if there is a cardinal between aleph_null and aleph_1, so aleph_null<cardinality of R which may or may not be = alepth_1.
I'm sure there are many people far more knowledgeable than me who can help you.
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my friends from oxford university said that aleph on is infinite^infinite while aleph one is beyond that also uncountable infinite = aleph one
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Aleph one is anything related to transinfinity, and thus, superior to also infinite^infinite.
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Aleph one would be uncountably infinite above aleph null even with having infinite amount of power sets for infinite sets would still not be possible. That's why you reach a certain ceiling called epsilon naught which is far beyond that of ω^ω^ω^ω and so on.
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To put in blunt, your friend is not incorrect assuming he was talking about ordinal number, infinite^infinite is nowhere as close as Aleph 1 in ordinals, but is Aleph 1 in cardinals which even 2^infinite is higher infinity.
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Peringatan 1x up post lama
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Buset.
