If $Y \subset O$ and $O$ is open, then $\overline Y \subset O$.

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?

• you are not so clear about your explanation for metric space thing.... what do you want to conclude after $d(\bar y, Y) = 0$?
– user87543
Feb 8, 2014 at 14:19
• $\subset$ is strict inclusion? Feb 8, 2014 at 14:19

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: $Y=(0,1),\ O=(0,2)$.

• I guess he mean proper containment... It is obvious that it is not true if we take $Y=O$
– user87543
Feb 8, 2014 at 14:18
• $Y=(0,1),\ O=(0,2).$ Feb 8, 2014 at 14:21
• This is fantastic.... you better make it as content of your answer.... So simple so perfect.... $Y=(0,1), O=(0,2)$
– user87543
Feb 8, 2014 at 14:22

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 $Y = \{1 - \frac{1}{n}\}_{n \in \mathbb{N}}$

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 $d(\bar y, Y) = 0$.