# Proving that ${\aleph_1}^{\aleph_0}\leq |[\omega_1]^{\omega}|$

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?

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

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.