Let $(x_n),(y_n)$ be two sequences in a metrizable topological space $X$ such that none of $(x_n),(y_n)$ has cluster point in $X.$ Can we then conclude the set $\{x_n:n\ge1\}\cup\{y_n:n\ge1\}$ has no accumulation point in $X?$
I think that should be correct yet I cannot get sure.
Please help.
Yes. Because if $z_0$ is an accumulation point for $A=\{x_n:n\ge1\}\cup\{y_n:n\ge1\}$ then there is a subsequence of $A$ like $(z_n)_{n\in \mathbb{N}} $ such that $z_n\to z_0$. But either the set $\{ z_n|n\in \mathbb{N}\} \cap \{x_n:n\ge 1\} $ have infinte elements or $\{ z_n|n\in \mathbb{N} \}\cap \{y_n:n\ge1\}$ have infinite elements. So there is a subsequence of $(z_n)$ like $(z_{n_k})_{k\in \mathbb{N}} $ such that $(z_{n_k}) \subseteq \{x_n:n\ge1\}$(or $\subseteq \{y_n:n\ge1\} $). But $z_{n_k}\to z_0$. So $z_0 $ is an accumulation point for $\{x_n:n\ge1\}$(or for $\{y_n:n\ge1\}$) and this is a contradiction.