the next smallest Aleph is defined by the successor operation, or basically the cardinality of the smallest Ordinal Number into which you can make an injective map, but cannot map it back into the set. Aleph-1 in this case is the smallest uncountable Infinite Cardinal greater than Aleph-0.
Thanks so 2 power aleph 0 ≠ aleph 1 ?
2^aleph0 = aleph 1.
It is both consistent that 2^Aleph-0 = Aleph-1 and 2^Aleph-0 = Aleph-n for any n > 0 (except some special cases like aleph-ω and cardinals with cofinality ω), the assertion that 2^Aleph-0 = Aleph-1 is what we call the Continuum Hypothesis, which can be further generalized to GCH (Generalized Continuum Hypothesis), which says that 2^Aleph-n = Aleph-(n+1), the truth of the Continuum Hypothesis is independent from ZF (Zermelo-Fraenkel Set theory, the most commonly used axiomatic system of set theory). So the answer to your question is that it cannot be proven in ZFC or ZF.
Nerd
Ratio.
So 2^aleph 1665 = aleph 1666 ?
If you assume GCH to be true, then yes. Though not necessarily.
As for our standards we assume continuum hypothesis to be true, so yes.
Thanks
I have a last question that will remove my remaining doubts (I know that cardinal = what constitutes the set) so my question would be the cardinal of aleph 0 would be greater than or equal to aleph 0?
Equal.
A cardinal number is a number which measures the cardinality of sets, the cardinality of a set is what you're talking about, and Aleph-0 = Aleph-0 of course. Though i think your question is more "If we look at Aleph-0 in terms of sets, then what is its cardinality?", and the answer is, still Aleph-0.
Now, let's address the accuracy of the statements made in the article:
Scans?
|R| = 2^ℵ0 so |R| = ℵ1Continuum hypothesis - Wikipedia
en.wikipedia.org
2^ℵ0 = ℵ1
The bolded part is incorrect, it can be proved that the cardinality of the power set of N is exactly equal to |R|, the rest is correct (Hint: to prove the bijection between P(N) and R, either consider the function f(A) = ∑2^{-r} for r ∈ A and A ⊆ N which pretty much constructs a bijection between P(N) and [0, 1], or construct an injection f: P(N) → R and another injection g: R → P(N), then use the Schröder–Bernstein theorem to prove a bijection).
I never said anything about additional axioms, i said that it cannot be proven nor disproven within ZF or ZFC.
The |R| = ℵ1 is redunant here and it isn't necessary to include it, since |P(N)| = 2^|N| = |R| which directly implies |R| = Aleph-1 assuming GCH.
I said, "Which means that the size of the power set of the set of natural numbers is strictly larger than the size of the set of real numbers."
Let me even correct my mistake using the Schröder–Bernstein theorem.
Eh... Not exactly superfluous. we explain the same thing in two different ways, just that's it.
You need to know what the mathematical symbols used in set theory's mathematical formulas and equations mean.
No problem at all.
The proof works, just a small correction, the functions should be denoted f: P(N) → R and g: R → P(N) since |R| denotes the cardinality of R while this is a function between sets.
You might have misunderstood what i said a little bit, i mean that it is redunant to include |R| = ℵ1 in your expression of GCH, because it is already included in the second part of your statement, which is also very slightly incorrect because the assumption λ < 2^λ has little to do with GCH and is already proved by Cantor's Theorem. I think what you meant to express is: ∀λ ∈ Card, ¬(∃κ : λ < κ < 2^λ). Which is also equivalent to: ∀α ∈ ORD, ℵ(α+1) = 2^(ℵα).
I suggest you read: https://en.wikipedia.org/wiki/Aleph_number
