By definition $\aleph_1 = 2 ^{\aleph_0}$. And since $2 < \aleph_0$, then $2^{\aleph_0} = {\aleph_1} \le \aleph_0 ^ {\aleph_0}$. However, I do not know what exactly $\aleph_0 ^ {\aleph_0}$ is or how I could compute it.
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10$2^{\aleph_0}$ is not the definition of $\aleph_1$. Do you know of the continuum hypothesis? (It is the definition of $\beth_0$ however.) – anon Aug 17 '12 at 05:47
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4See e.g. Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$? You can find pointers to some basic questions on cardinals, which already have been asked (and answered) at MSE, here: Overview of basic results on cardinal arithmetic. – Martin Sleziak Aug 17 '12 at 05:51
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4@anon: You mean $\beth_1$, since $\beth_0=\aleph_0$... – Asaf Karagila Aug 17 '12 at 05:55
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@AsafKaragila Oops. – anon Aug 17 '12 at 05:57
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No. By definition $\aleph_1$ is the least uncountable $\aleph$ number. $2^{\aleph_0}$ can be quite a large $\aleph$, or it could be $\aleph_1$. For example, many forcing axioms (e.g. the proper forcing axiom) prove that $2^{\aleph_0}=\aleph_2$.
The assertion $2^{\aleph_0}=\aleph_1$ is known as The Continuum Hypothesis and was proven unprovable from the usual axioms of set theory. We can therefore add axioms which decide the continuum hypothesis, e.g. itself or the aforementioned forcing axiom.
On the the other hand:
$$2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}= 2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$
To read more:
Here are some links to answers discussing the cardinality of the continuum:
Asaf Karagila
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