I recently learned that in the metric space $(0, 1)$ w.r.t. the usual metric $d$, $(0, 1)$ is closed in $((0, 1), d)$. This means that $\overline{(0, 1)}$ = $(0, 1)$ in $((0, 1), d)$.
Now, if we take the sequence $(\frac{1}{n})_{n \in \mathbb{N}}$ in $(0, 1)$, then it converges to $0$ w.r.t. $d$. This means $0$ is an accumulation point. Does this necessarily mean $0 \in \overline{(0, 1)}$? $0$ is an accumulation point, so it's in $(0, 1)'$, but I'm not sure if it is in the closure of $(0, 1)$. I don't think it is, since $(0, 1)' \ne \overline{(0, 1)}$, but I'm not sure if that's the correct way to think about it.
If indeed $0 \in \overline{(0, 1)}$ in $((0, 1), d)$, that does not imply $0 \in (0, 1)$, right, the the "union" definition of closure above?
By definition, $\overline{(0,1)}$ is the smallest closed set in $(0,1)$ that contains $(0,1)$, which is $(0,1)$. The point $0$ is not a point in the metric space $\big((0,1),d\big)$. The sequence $(1/n)$ does not converge to anything in this metric space.