Is $\\aleph_0^{\\aleph_0}$ smaller than or equal to $2^{\\aleph_0}$?
We need to find an injection from $\mathbb N^\mathbb N$ to $2^\mathbb N$. We need to encipher a sequence of natural numbers as a binary sequence.
elementary set theory - How much is ${\aleph_0}^{\aleph _ 0 ...
So the strategy for proving equalities of cardinals is always to find an upper and a lower bound for them. Now let us do this in your example: $$2^ {\aleph_0}\le \aleph_0^ {\aleph_0} \le \left (2^ {\aleph_0}\right)^ {\aleph_0} = 2^ {\aleph_0\cdot \aleph_0} = 2^ {\aleph_0}$$. Therefore Cantor-Bernstein implies $\aleph_0^ {\aleph_0}=2^ {\aleph_0 ...
What is $\\aleph_0!$? - Mathematics Stack Exchange
2017年12月5日 · ℵ 0! is the cardinality of the set of bijections N →N N → N, and it's not hard to show that this has cardinality ℵℵ00 ℵ 0 ℵ 0, for example using Cantor-Bernstein-Schroder. We have k! = ⋅ ⋯ k k! = ⋅ ⋯ k, i.e., it is products of all numbers with size at most k k. Therefore.
What are $\aleph_0$, $\omega$ and $\mathbb {N}$ and how are …
2020年4月18日 · $\aleph_0$ is the smallest infinite cardinal, i.e. the cardinality of $\mathbb{N}$. Under the von Neumann cardinal assignment, $\aleph_0$ is defined as $\omega_0$. However, using the symbol $\omega_0$ or $\aleph_0$ indicates whether we are treating it as an ordinal or as a cardinal, respectively.
Cardinality of $\\aleph _0$ - Mathematics Stack Exchange
2019年3月25日 · 2. Part 1 1: The cardinality is uncountable, with a caveat: The cardinality of this equivalence class is, at first glance, uncountable. For notation's sake, let us denote this class by ℵ0¯ ¯¯¯¯ ℵ 0 ¯, to avoid confusion with the cardinal. It is known that the cardinality of the reals - that of the continuum - is greater than that of ...
Questions about Aleph-Aleph-Null - Mathematics Stack Exchange
2019年1月2日 · Yes, $\aleph_0=\omega$, but we write $\omega$ when we think of it as an ordinal instead of a cardinal. The rest of your question is based on a fundamental misunderstanding: $\aleph_ {\alpha+1}$ is not the cardinality of the power set of $\aleph_\alpha$; what it is is the smallest cardinal larger than $\aleph_\alpha$.
$\aleph_1 = 2^ {\aleph_0}$ - Mathematics Stack Exchange
By Cantor's diagonal argument, $2^ {\aleph_0}$ is uncountable. Meanwhile, $\aleph_1$ is defined to be the cardinality of the smallest uncountable ordinal; equivalently, it is the number of distinct ordertypes of countable ordinals. By an argument similar to that of the Burali-Forti paradox, $\aleph_1$ is uncountable, and (assuming a very mild ...
Understanding the proof for $ 2^ {\aleph_0} > \aleph_0$
Added You can also prove that $2^{\aleph_0} > \aleph_0$ by showing that $2^{\aleph_0}$ is the cardinality of $[0,1)$.
elementary set theory - Is the logarithm of $\aleph_0$ infinite ...
2015年5月7日 · You note that saying here $\log_2 \aleph_0=\aleph_0$ is in contradiction with André Souza Lemos' answer, and $2^{\aleph_0}=\aleph_0$ contradicts the assertion made in the question that $2^{\aleph_0}=\aleph_1$. But this is only because I have been considering a different mathematical context.
Can you go from $\\aleph_0$ to $\\aleph_1$ with tetration or …
2016年2月2日 · $\begingroup$ Since $2^{\aleph_0} = {\aleph_0}^{\aleph_0}$, we could write the expression out as $\aleph_0 \uparrow\uparrow 2$, and phrase the Continuum Hypothesis in those terms, but unless there's some route through tetration to "proving" the CH, the corresponding principle would still be hypothetical. $\endgroup$