Closure in topology space 
*

*We have the set $X$. $X=\{1,2,3,4\}$.


*$\tau$- topology in $X$, $\tau =\{\emptyset, \{1,2,3,4\},\{1,2\},\{2,3\},\{1,2,3\}, \{2\}\}.$


*$(X,\tau )$ - space.
$A=\{1,2\}$. Find the closure $\bar{A}$ of $A$ in $(X,\tau )$.
I have found  that $\bar{A}=\{1,2,3,4\}.$
Is it correct? It is my own example. I just want to check my understanding  of this concept. I get this answer by looking at the pictire of this topology.
Are there more time efficient methods for doing this? Maybe some theorem or trick or something like this?
Edit:  the definition from my textbook.

 A: You are right. One thing to note is that $\overline{A}$ always contains A. It is the smallest closed set containing A. So in that sense, $\{1, 2, 3, 4\}$ is the smallest closed set in your space containing $A$.
A: Your definition of closure means that the complement of $\overline{A}$ is precisely the union of all open sets that don’t meet $A$:
$$X\setminus\overline{A} = \bigcup\{U : U\in \tau, A\cap U =\varnothing\}$$
Now $A\cap U=\varnothing$ is equivalent to $A\subset X\setminus U$.
Thus, by taking complements, we see that $\overline{A}$ is the intersection of all closed sets that contain $A$:
$$\overline{A} = \bigcap\{X\setminus U : U\in\tau, A\subset X\setminus U\}$$
(This is true in general and is often the most useful characterization of $\overline{A}$.)
In your case, you can explicitly list the small collection of closed sets $X\setminus U$. Only the closed set $\{1,2,3,4\}$ contains $A$, so this is the only set in the intersection.
A: The closure is correct, we add $4$ because $4$ an open set that contains $4$ must be $X$ (just check the list) and thus intersects $A$ as $A$ is non-empty.
An open set that contains $3$ also contains ${2,3}$ and thus intersects $A$ too. So besides $A$ itself we get $3,4$ as points of the closure too. So $\overline{A} = \{1,2,3,4\}$ indeed.
