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Let $X$ be a topological space, $Y \subset X$. Is it true that if $Y \subset O$ and $O$ is open, then $\overline Y \subset O$?
If $X$ is a metric space it is true since if $\bar y \in \overline Y \setminus Y$ then $d(\bar y, Y) = 0$. Does the result still holds for arbitrary topological spaces? If not, what is the essential request to make it work?
Isn't. Take $Y=O$ in ${\Bbb R}$ with the usual topology. And in an arbitrary topological space there isn't distance.
With strict inclusion: $Y=(0,1),\ O=(0,2)$.
In general, it can fail pretty badly.
If the space is irreducible for example, then any containment $X \subset Y$ with $X$ non-empty and open and $Y$ not being the entire space will be a counter example since then $\overline{X}$ is the entire space.
If $Y \neq O$, take $O = D(0,1) $ and $Y = \{1 - \frac{1}{n}\}_{n \in \mathbb{N}}$
The answer to your second question is yes.
If $\bar y \in \overline Y$, then for every $\epsilon > 0$ there exists $y \in Y$ such that $d(\bar y, y) < \epsilon$. Thus, for every $\epsilon > 0$, $d(\bar y, Y) < \epsilon$, which is equivalent to $d(\bar y, Y) = 0$.
In a generic topological space there is no notion of distance.
