Topology without tears exercises 1.2 #6 i) Let T be a topology on a set X such that T consists of precisely four sets; that is , $T = \{X, \emptyset, A, B\}$, where $A$ and $B$ are non empty distinct proper subsets of $X$. Prove that $A$ and $B$ must satisfy exactly one of the following conditions: 
a) $B = X\setminus A$
b) $A \subset B$
c) $B \subset A$
hint: show that A and B must satisfy at least one of the conditions and thens show they cannot satisfy more than one of the conditions.
Attempt: So I tried this, but it doesn't seem to complete in my opinion.
Assume a) i.e $B = X\setminus A$:
Then $A\cup B = X \in T$ and $A \cap B = \emptyset \in T$
So I established that it satisfies the topological conditions, but now to show that the other conditions cannot be satisfied at the same time how could I go about it?  I mean if I explain it in just words it would be that since I assumed condition a) then there is no way for b) or c) to even be valid. But there has to be a way to show it no?
 A: It may be helpful to break it down into several claims to be proved:


*

*$\boxed{\text{(a) and (b) cannot both hold}}:$ Suppose they did. Since $A \neq \varnothing$, there is some $a \in A$. But $A \subseteq B$, so $a \in B$. But then $a \in A \cap B = \varnothing$, a contradiction.

*$\boxed{\text{(a) and (c) cannot both hold}}:$ Similar reasoning as above.

*$\boxed{\text{(b) and (c) cannot both hold}}:$ Suppose they did. Then $A = B$, contradicting the assumption that all four sets were distinct.

*$\boxed{\text{At least one of (a), (b), and (c) must hold}}:$ If (b) or (c) hold, then we are done. So we may assume that $A \not\subseteq B$ and $B \not\subseteq A$. We want to show that (a) holds.
Now consider $A \cup B$, which by definition of a topology must be either $\varnothing$, $A$, $B$, or $X$. Since both $A$ and $B$ are nonempty, we know that $A \cup B \neq \varnothing$. Since $B \not\subseteq A$, we know that $A \cup B \neq A$. Since $A \not\subseteq B$, we know that $A \cup B \neq B$. So $\boxed{A \cup B = X}$.
Likewise consider $A \cap B$, which by definition of a topology must be either $\varnothing$, $A$, $B$, or $X$. Since $A$ and $B$ are proper subsets of $X$, we know that $A \cap B \neq X$. Since $B \not\subseteq A$, we know that $A \cap B \neq B$. Since $A \not\subseteq B$, we know that $A \cap B \neq A$. So $\boxed{A \cap B = \varnothing}$.
Combining the last two boxed equations, we conclude that (a) holds, as desired. $~~\blacksquare$
A: Hints:


*

*$A\cup B\in T$ since $T$ is assumed to be a topology, you have
for possibilities that one of which must hold - for example if $A\cup B=X$
then $B=X\setminus A$, what if $A\cup B=A$ or $A\cup B=B$ ? can
$A\cup B=\emptyset$ ?

*Now, can both of those three options hold simultaneously ? for
example if both (b) and (c) are true then $A=B$, but you know $A\neq B$ 
