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Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{rank}_{CB}(\text{ran}(\varphi_\alpha)) = 1$, i.e. $\text{ran}(\varphi_\alpha)$ is made of isolated points, for all $\alpha\in \omega_1$.

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}_{\omega_1}$.
My questions are:

1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega_1$ subset of the reals"? In case it is not, is it consistent $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})+$ The above statement $+$ "It does not exists an $\omega_1$ subset of the reals"?

Thanks!

Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{ran}(\varphi_\alpha)$ is closed and $\text{rank}_{CB}(\text{ran}(\varphi_\alpha)) = 1$, i.e. $\text{ran}(\varphi_\alpha)$ is made of isolated points, for all $\alpha\in \omega_1$.

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}_{\omega_1}$.
My questions are:

1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega_1$ subset of the reals"? In case it is not, is it consistent $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})+$ The above statement $+$ "It does not exists an $\omega_1$ subset of the reals"?

Thanks!

Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{rank}_{CB}(\text{ran}(\varphi_\alpha)) = 1$, i.e. $\text{ran}(\varphi_\alpha)$ is made of isolated points, for all $\alpha\in \omega_1$.

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}_{\omega_1}$.
My questions are:

1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega_1$ subset of the reals"?

Thanks!

Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{rank}_{CB}(\text{ran}(\varphi_\alpha)) = 1$, i.e. $\text{ran}(\varphi_\alpha)$ is made of isolated points, for all $\alpha\in \omega_1$.

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}_{\omega_1}$.
My questions are:

1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega_1$ subset of the reals"? In case it is not, is it consistent $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})+$ The above statement $+$ "It does not exists an $\omega_1$ subset of the reals"?

Thanks!

Consider the following statement (in $\mathsf{ZF}$):

There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{rank}_{CB}(\text{ran}(\varphi_\alpha)) = 1$, i.e. $\text{ran}(\varphi_\alpha)$ is made of isolated points, for all $\alpha\in \omega_1$.

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}_{\omega_1}$.
My questions are:

1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega_1$ subset of the reals"?

Thanks!

Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{rank}_{CB}(\text{ran}(\varphi_\alpha)) = 1$, i.e. $\text{ran}(\varphi_\alpha)$ is made of isolated points, for all $\alpha\in \omega_1$.

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}_{\omega_1}$.
My questions are:

1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega_1$ subset of the reals"?

Thanks!