In topological spaces are there at least two equivalent ways to define the closure of a set?

Question

Let $(X, F)$ be an arbitrary topological space and $Y$ be a subset of $X$.
Are the following sets always the same?
$Y_1$ $:=$ the set of all adherent point of $Y$.
$Y_2$ $:=$ the intersection of all closed subsets of X containing $Y$.

Answer 1

Yes these are always the same.

Suppose $Y_1$ is the set of adherent points of $Y$. If $x \notin Y_1$, there is an open set $O_x$ that contains $x$ and that does not intersect $Y$ (this is what the negation of being an adherence point comes down to). Note that all points of $O_x$ are in fact not an adherence point of $Y$ as well: the same $O_x$ works for them too. So $O_x \subseteq X\setminus Y_1$.

So $\bigcup \{O_x: x \in X\setminus Y_1\} = X\setminus Y_1$ (we just showed the left to right inclusion, the other one is obvious), and so $X\setminus Y_1$ is open (as a union of the open sets $O_x$) and so $Y_1$ is closed.

So $Y_2 \subseteq Y_1$ as $Y_1$ is one of the closed sets that contain $Y$ (as we just saw) that we are taking the intersection of to define $Y_2$.

In the other hand, if $x \notin Y_2$, this means there is some closed set $F$ with $Y \subseteq F$ so that $x \notin F$ (this is what not being in the intersection defining $Y_2$ means). But then $X \setminus F$ is an open set that contains $x$ and that misses $Y$ (as $Y \subseteq F$, $X\setminus F \subseteq X\setminus Y$) and so $x$ is not an adherent point of $Y$. So $X\setminus Y_2 \subseteq X\setminus Y_1$ or $Y_1 \subseteq Y_2$.

Both inclusions together imply equality.

Answer 2

In my opinion one should use $$\overline{Y} = \bigcap_{C \subset X \text{ closed with } Y \subset C} C$$ as the definition of the closure. This makes clear that $\overline{Y}$ is the smallest closed set containing $Y$.

Then

$\overline{Y}$ = set of all adherent points of $Y$

can be regarded as a theorem (which was proved in Henno Brandsma's answer), but if you absolutely want you can also use it as an alternative definition.