Determine whether a set is closed? Let $X=\{1,2,3\}$ and Topology $T=\{\{1,2,3\}, \{1,2\}, \{3\},\emptyset\}$.
If $A=\{1\}$ and $B=\{1,2\}$. Are they closed?
I am stuck on this question. How can we find the complement of the given $X$? Shouldn't there be more information?
 A: Recall that the definition of a closed set $U\subset X$ is a subset such that $X\setminus U$ is an open set (that is $X\setminus U\in\tau$).
Now, for $A=\{1\}$ we have $X\setminus A=\{2,3\}$ and for $B=\{1,2\}$ we have $X\setminus B=\{3\}$. Can you see these sets in the topology $\tau=\{X,\{1,2\},\{3\},\emptyset\}$?
A: The complement of the subset $A$ is every element in $X$ which is not in $A$. So if you know the elements of $X$ and the elements $A$, you have enough information to find the complement. What points in $X$ are not in $A$? Clearly the answer is $2$ and $3$. So the complement of $A$ is $\{2, 3\}$. Is this set open? Well it's not a subset of the topology on $X$ so by definition the answer is no. Just use this same process to determine if $B$ is closed.
A: Your information here is a bit sketchy.  I believe that $X$ here is the listing of open sets in the topology you are considering.  (In pointset topology, one describes the topology by explicitly listing the open sets.  The only thing that troubles me is that the empty set should have been included as a member of the set $X$, and I don't see it there.)  
Anyway, if I interpret things correctly, this is a topology with pointset $S=\{1,2,3\}$ whose open sets are the members of $X$ (although, as I said above, $\emptyset$ is somehow missing).  A closed set is a set that is the complement of n open set, where complementation is always performed relative to the pointset ($S=\{1,2,3\}$ in this case).  Thus the complement of $\{1\}$ would be $\{2,3\}$.  
As $\{2,3\}$ is not a member of $X$, it is not open.  Thus $A=\{1\}$ is not closed.  On the other hand, $B=\{1,2\}$ is closed.  (Can you see why?)   
