Let $(X,d)$ be a metric space and $(x_n)$ be a sequence in $X.$ If $(x_n)$ has no cluster point in $X$ then prove/disprove: $A=\{x_n:n\in\mathbb N\}$ is a closed and discrete subset of $X$ w.r.t. the metric induced from $d.$
I think the statement is correct. Since $(x_n)$ does not cluster in $X$, all the given $x_n$'s are isolated points of $A$ and hence $A$ is discrete.
I am not sure about the closed part. Can you help me?
So, let us prove that $A$ is closed. A definition of closed subset $S$ in metric space $X$ is that $S$ contains all of its limit points, i.e. any sequence $\{x_n\}\subset S$ convergent in $X$ is actually convergent in $S$. Now, if $A$ is a discrete set and $x_n $ is a sequence in $A$, then it can be convergent in $X$ only if it is constant from one index $\overline n$ on. Then the limit value is $x_{\overline n}\in A$.
Since $(x_n)_n$ has no cluster point all subsequences of $(x_n)_n$ do not converge. Moreover, in a metric space a set $A$ is closed iff for every convergent sequence in $A$ the limit is in $A$. This is trivially true in your example because the only convergent sequences are eventually constant (otherwise they would be subsequences of $(x_n)_n$).
