Cardinality of all cardinalities

Allow $C = \{0, 1, 2, \ldots, \aleph_0, \aleph_1, \aleph_2, \ldots\}$. What is $\left|C\right|$? Or is it also distinct?

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2019-05-05 03:14:07
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Answers: 4

I might be incorrect yet it appears that C is not a set of ready or anything comparable. It has IMHO an unimportant bijection right into $\mathbb{Z}$ and also consequently right into $\mathbb{N}$.

I have not fulfilled any kind of mystery to restrict $|C| \in C$ ($\{1\}$ is such set).

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2019-05-08 21:55:03
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C is not a set - it is infact a correct class. If C were a set, after that |C | would certainly be specified. It after that adheres to that |C | would certainly be the biggest cardinality, given that there is a complete order in between all the cardinalities, and also $|C| > \kappa$ for every single cardinality $\kappa$ (every cardinality amounts the set of all smaller sized cardinalities). Yet $2^{|C|} > |C|$ therefore there can not be a biggest cardinal.

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2019-05-08 08:18:30
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The most usual means to specify the principal number $|X|$ of a set $X$ is as the least ordinal which remains in bijection with $X$. After that $C$ is a boundless class of ordinals, and also any kind of such is always a correct class. Given that $C$ is not a set, it does not have a cardinality.

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2019-05-08 08:14:44
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This is "Fact 20" on web page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

These are notes on boundless collections from one of the most "ignorant" viewpoint (as an example, among the realities is that every boundless set has a countable part, so specialists will certainly see that some weak kind of the Axiom of Choice is being thought without comment. Yet this follows the means collections are made use of in mainstream maths). It is suggested to be obtainable to undergrads. Specifically, ordinals are not stated, although there are some more records - - change "1" in the link over with "2", "3" or "4" - - which define such points a little bit.

Yet I do not see why it is essential or handy to mention ordinals (or cosmos!) to address this inquiry.

Included : to be clear, I desire to modify the inquiry in the list below means :

There is no set $C$ such that for every single set $X$, there exists $Y \in C$ and also a bijection from $X$ to $Y$.

This is very easy to confirm using Cantor's diagonalization and also it avoids the "embodiment trouble for cardinalities", i.e., we do not require to claim what a cardinality of a set is , just to recognize when 2 collections have the very same cardinality. I think this is ideal for a basic mathematical target market.

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2019-05-08 07:21:00
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