We have symbols of cardinal numbers. The most known are aleph zero and continuum. Somewhere I've noticed the sequence of cardinal numbers as aleph zero, aleph one, aleph two... where $\aleph_n$ = $2^{\aleph_{n-1}}$. Can we say that aleph one = continuum?
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1Not in ZFC. – Feb 08 '14 at 23:54
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3That $\aleph_{n+1}=2^{\aleph_n}$ is the Generalized Continuum Hypothesis. See http://math.stackexchange.com/questions/666469/are-there-any-infinites-not-from-a-powerset-of-the-natural-numbers. – Martín-Blas Pérez Pinilla Feb 09 '14 at 00:00
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2@Martín-BlasPérezPinilla Your comment is correct if $n$ is understood to range over all ordinal numbers (so that it applies to exponentiation of all cardinals), but I fear that Emin will misunderstand $n$ as ranging only over the natural numbers. – Andreas Blass Feb 09 '14 at 01:23
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True. This explains the $n,n-1$. – Martín-Blas Pérez Pinilla Feb 09 '14 at 07:00
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Your notation is off by a letter (of the Hebrew alphabet).
The $\aleph$ numbers are not defined by taking power sets of the previous one, but rather increasing "just enough" the cardinality. What does that mean? It means that $\aleph_1$ is not defined as $2^{\aleph_0}$, but rather as the smallest possible uncountable cardinal. $\aleph_2$ is not defined as $2^{\aleph_1}$, but rather as the least possible cardinal which is larger than $\aleph_1$. And so on.
On the other hand, there are $\beth$ (Beth) numbers, which are defined by taking power sets. So $\beth_0$ is $\aleph_0$, and $\beth_1=2^{\beth_0}=2^{\aleph_0}$, and $\beth_2=2^{\beth_1}=2^{2^{\aleph_0}}$, and so on.
In both the $\aleph$ and $\beth$ definitions, after we exhaust the natural numbers as indices, we take limits, and repeat the construction. This eventually exhausts the entire class of ordinals.
From the usual axioms of set theory we cannot prove nor disprove that $\aleph_1=\beth_1$. So while we cannot say that $2^{\aleph_0}=\aleph_1$, we can certainly say that $2^{\aleph_0}=\beth_1$.
Asaf Karagila
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