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I have to prove that there are $2^{\aleph_0}$ injections from $\omega$ to $\omega_1.$ I can see that there is a bijection between this set and the set of pairs: (permutation of $\omega$, infinitely countable subset of $\omega_1$), which means that the standard "finite" formula works, and there are $$\aleph_0!\cdot{\aleph_1 \choose\aleph_0}$$

of such injections. I know that $\aleph_0!=2^{\aleph_0},$ but I can't see how the other factor is $2^{\aleph_0}$ too.

Bartek
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  • The usual notation is $[\omega_1]^\omega$ for countably infinite subsets of $\omega_1$. – Asaf Karagila Apr 20 '13 at 21:16
  • Also see http://math.stackexchange.com/questions/311848/proving-that-for-infinite-kappa-kappa-lambda-kappa-lambda and also http://math.stackexchange.com/questions/319145/cardinality-of-a-set-containing-subsets-of-omega-1 and also http://math.stackexchange.com/questions/312431/what-is-the-value-of-aleph-1-aleph-0 – Asaf Karagila Apr 20 '13 at 21:16
  • I prefer $\mathcal P_{\omega_1}(\omega_1)$ or $[\omega_1]^{\aleph_0}$. Typically, I use $[\omega_1]^\alpha$ for the set of increasing $\alpha$-sequences of countable ordinals. – Andrés E. Caicedo Apr 21 '13 at 04:57

1 Answers1

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Since every countable subset of $\aleph_1$ is bounded, you can get a handle on it as follows:

$${\aleph_1 \choose\aleph_0} = \bigcup_{\alpha < \omega_1}{\alpha \choose\aleph_0}$$

Amit Kumar Gupta
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