Here is the question:
Let $X = \mathbb{R}$ and let $\Omega$ consist of the empty set and all infinite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?
My attempt : I think the answer is No; it is not a topology.
Because we have $$ \Omega = \lbrace \varnothing \rbrace \cup \lbrace U \subseteq \mathbb{R} \mid \text{$U$ is infinite} \rbrace $$
If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $\Omega$ now $\cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $\Omega$. Is this explanation correct?
EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $\Omega$ as they are infinite sets but $ U_1 \cap U_2 = \lbrace 2 \rbrace $ which is finite.
Best Answer
No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is $\{2\}$). This violates the condition that the intersection of two open sets be open in a topology.
However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$\Omega=\{\varnothing\}\cup\{U\subseteq R\,|\,U^{\mathsf c}\text{ is finite}\}.$$