Is this Topology Proof Rigorous? Question:
On the three-point set X = {a, b, c}, the trivial topology has two open sets and the discrete topology has eight open sets. For each of n = 3,...,7, either find a topology on X consisting of n open sets or prove that no such topology exists.
I was able to find a topology for each n except n = 7. So I tried to prove that the topology does not exist this way:
We prove this by contradiction. Suppose n(T) = 7 exists for a three-point set X, then the elements of T will either be:
(i) the empty set, X, two singleton sets and three sets each of two elements or
(ii) the empty set, X, three singleton sets and two sets each of two elements
case(i): this is impossible since the intersection of elements of T will not always be closed (there exist a pair of sets each of two elements whose intersection is a singleton set not in T)
case(ii): this is impossible since the union of two singleton sets is not always closed (there exist a pair of singleton sets whose union will give a set with two elements not found in T).
Since we have shown that both possibilities of the elements of T do not exist, then n(T) = 7 is not a topology for a three-point set X and this concludes the proof.
Note: I feel this proof is not rigorous. I will welcome contributions from you guys. I will also like to see alternative proof. Thanks
 A: Suppose that $\tau$ is a topology on $X=\{1,2,3\}$ such that $|\tau|=7$, which means there is a unique set $A \subseteq X$ such that $A \notin \tau$, as $X$ has indeed exactly $8$ subsets. 
$|A| = 0$ is impossible as then $A=\emptyset$ and $\emptyset \in \tau$ by one of the topology axioms.
$|A|=3$ is also impossible as then $A=X$ and $X \in \tau$ by axiom.
If $|A|=1$ we have $A=\{i\}$ for some $ i \in X$ but then let $j,k$ be the two other elements of $X$. We must have $\{i,j\}, \{i,k\} \in \tau$ as $A$ was the unique subset not in $\tau$ and so these must be in $\tau$. But then $\{i,j\}\cap \{i,k\} = \{i\} \in \tau$ too, as a finite intersection of members of $\tau$. This is a contradiction. 
The final case is $|A|=2$ so $A= \{i,j\}$ for $i,j \in X$. But again we must have that $\{i\},\{j\} \in \tau$ (other subsets are in $\tau$) and then $\{i\} \cup \{j\}=A \in \tau$ (closedness under unions), contradiction. So all cases lead to contradictions and we've shown the impossibility of $|\tau|=7$.
