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Today I was working with some exercises about topology but in some part I need to prove the next inequality: $${\aleph_1}^{\aleph_0}\leq |[\omega_1]^{\omega}|$$Here $[\omega_1]^{\omega}:=\left\{A\subseteq\omega_1 : |A|=\aleph_0 \right\}$. I don't know how to prove it.

My attempt starts with $f:\omega\to\omega_1$ a function. We know that ${\aleph_1}^{\aleph_0}=|\left\{f:\omega\to\omega_1\mid f \ \text{is a function} \right\} |$ then is correct to take $f:\omega\to\omega_1$. Then $f[\omega]$ is a subset of $\omega_1$ but then $f:\omega\to f[\omega]$ is a surjective function. Then $\aleph_0\geq |f[\omega]|>0$ and therefore every function from $\omega\to\omega_1$ defines naturally a subset of $\omega_1$. The problem is the fact that $f[\omega]$ can be a finite set and the natural asignation $f\mapsto f[\omega]$ doesn't works to prove the inequality. Moreover, I think that this asignation isn't inyective because we can have two different functions $f$ and $g$ such that $f[\omega]=g[\omega]$. For example $f(n)=n$ and $g(0)=1$, $g(1)=0$ and $g(n)=n$ for $n>1$. I don't know how to procced or a way to conclude the exercise. Anyone can help me?

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    $\begingroup$ Isn't the set of all functions from $\omega$ to $\omega_1$ a subset of $[\omega\times\omega_1]^\omega$? $\endgroup$
    – bof
    Commented Jun 12, 2020 at 0:11
  • $\begingroup$ I suppose you know $|\omega\times\omega_1|=\aleph_0\cdot\aleph_1=\aleph_1$. Anyway, $(n,\xi)\mapsto\omega\xi+n$. $\endgroup$
    – bof
    Commented Jun 12, 2020 at 0:18

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You know that $2< \aleph_{1}\leq 2^{\omega}$. Then you have $2^{\omega} \leq \aleph_{1}^{\omega}\leq (2^{\omega})^{\omega}= 2^{\omega \times \omega}= 2^{\omega}$. It is clear that $ 2^{\omega}\leq|[\omega_{1}]^{\omega}|$ because you know that $|P_{\infty}(\omega)|= 2^{\omega}$ and $P_{\infty}(\omega)\subset [\omega_{1}]^{\omega}$, where $P_{\infty}(\omega)=\{A\subset \omega:A$ is infinite $\}$. Therefore you have your inequality, moreover you have the equality.

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