Set of infinite subsets. Is it a topology? 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.
 A: 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: $$\Omega=\{\varnothing\}\cup\{U\subseteq R\,|\,U^{\mathsf c}\text{ is finite}\}.$$
A: You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just $\{1\}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.
A: I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $\mathbb{R}$ But it does not have $\mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $\mathbb{R}$ belong to taa but it have all infinite subsets of $\mathbb{R}$,  not $\mathbb{R}$ itself. That's why in my view's taa is not a topology 
A: Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..
