Let $\kappa$ be a cardinal number, define $\textsf{M}(\kappa)$ and $\textsf{BW}(\kappa)$ as follows:
$\textsf{M}(\kappa)$ : For every sequence $(f_{n}:\kappa\to \mathbb{R})_{n\in\mathbb{N}}$ of real-valued function defined on $\kappa$ there is a subsequence $(f_{n}:\kappa\to \mathbb{R})_{n\in A}$ such that for each $\alpha\in\kappa$ the sequence $(f_{n}(\alpha))_{n\in A}$ is eventually monotonic.
$\textsf{BW}(\kappa)$ : For every sequence $(f_{n}:\kappa\to \mathbb{R})_{n\in\mathbb{N}}$ of pointwise bounded real-valued functions defined on $\kappa$ there is a subsequence $(f_{n}:\kappa\to \mathbb{R})_{n\in A}$ which is pointwise convergent.
Proposition: For each cardinal number $\kappa$, we have that $\textsf{M}(\kappa) \Longrightarrow \textsf{BW}(\kappa)$
Proof: Let $(f_{n}:\kappa\to \mathbb{R})_{n\in\mathbb{N}}$ be a sequence of pointwise bounded real-valued functions defined on $\kappa$ (i.e., for each $n\in\mathbb{N}$, the set $\{f_{n}(\alpha):\alpha\in\kappa\}$ is bounded on $\mathbb{R}$). As $\textsf{M}(\kappa)$ holds, there exists a subsequence $(f_{n}:\kappa\to \mathbb{R})_{n\in A}$ such that for each $\alpha\in\kappa$ the sequence $(f_{n}(\alpha))_{n\in A}$ is eventually monotonic. Then, for each $\alpha\in\kappa$, there is an infinite subset $A_{\alpha}$ of $A$ such that $(f_{n}:\kappa\to \mathbb{R})_{n\in A_{\alpha}}$ is monotonic, consider $A^{\prime}=\bigcap_{\alpha<\kappa}A_{\alpha}$.
Can I conclude that $A^{\prime}$ is infinite and that $(f_{n}(\alpha))_{n\in A^{\prime}}$ is pointwise convergent (in fact, only I need that $(f_{n}(\alpha))_{n\in A^{\prime}}$ is bounded)? In case my reasoning is wrong, can someone give me an idea of how to fix it?
Thanks a lot
