Complement of all finite subsets of $\mathbb{R}$

I was skimming through the topology book recommended through this question and I came across a question that I apparently solved incorrectly. It's question 2.5 on page 12.

Repeated here:

Let $X$ be $\mathbb{R}$, and let $\Omega$ consists [sic] of the empty set and complements of all finite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?

And my answer is "no", because I can posit some finite intersection of infinite sets and come up with a finite set, which is no longer in $\Omega$. For example, for some $a,b \in \mathbb{R}$ and $a<b$, two infinite sets are $X_1 = (-\infty,b)$ and $X_2 = (a,\infty)$ and obviously, $X_1, X_2 \in \Omega$ but $X_1 \cap X_2 \notin \Omega$.

But it seems it is a topology, because in the line following the question, they state:

The space of Problem 2.5 is denoted by $\mathbb{R}_{T_1}$ and called the line with $T_1$-topology.

So where am I going wrong?

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Your $X_1$ and $X_2$ are not the complements of finite sets, so they're not in $\Omega$. –  Chris Eagle Mar 17 '11 at 23:19
the finite complement topology is often used as an example of a non-hausdorf topology, as any two non-empty open sets intersect. –  yoyo Mar 17 '11 at 23:38

The problem with your argument is that not all infinite subsets of $\mathbb R$ are complements of finite sets. To see that we in fact have a topology, let $X_1$ and $X_2$ be finite subsets of $\mathbb R$ and observe that $(\mathbb R - X_1) \cap (\mathbb R - X_2) = \mathbb R - (X_1 \cup X_2)$ which is again the complement of a finite set. The union property should be obvious.
Not all infinite subsets of $\mathbb{R}$ are complements of finite subsets of $\mathbb{R}$. For example, $X_1=(-\infty,b)$ is the complement of $[b,\infty)$, which is infinite. Thus $X_1$ is not open in this topology.