$\mathbb{R}$ with the finite complement topology Let $X=\mathbb{R}$ be given with the collection $\tau$ where 
$$
\tau = \{U\subset X: |X\setminus U|&lt;\aleph_0\}\cup \{\emptyset\}
$$
I was sitting at my computer, when I suddenly asked myself:
"How is $\tau$ actually a topology on $\mathbb{R}$?". It is supposed to satisfy the arbitrary union condition, but it doesn't. Indeed
$$
\bigcup_{i\in \mathbb{N}} \mathbb{R}\setminus \{i\} \notin \tau
$$
but each of $\mathbb{R}\setminus \{i\} \in \tau$.
Can someone explain to me how $\tau$ is a topology?

edit
as Thomas Andrews points out, I was confusing unions with intersections and, in general, just being an idiot.
To get a topology you need to include $\emptyset$ in $\tau$. With that fixed the proof is easy.


$\tau$ is closed under intersection of two sets (and hence finitely many). If $A$ and $B$ are both in $\tau$ and neither is empty, then there are only finitely many elements missing from $A$ and finitely many missing from $B$. For an element $x$ to not be in $A\cap B$ we must have either $x\notin A$ or $x\notin B$, which can happen only for finitely many $x$. Of course, if one of $A,B$ is empty, so is the intersection so $A\cap B$ is still in $\tau$
Closure under unions is even easier. If $U_i$ are elements of $\tau$ for some indices $i\in I$, then if at least one of the sets, say $U_{i_1}$, is non-empty then there are only finitely many elements of $X$ missing from $U_{i_1}$. For an element $x\in X$ to not be in $\bigcup_{i\in I}U_i$ we must have $x\notin U_{i_1}$, so this can only happen for finitely many $x\in X$. Again, if all the $U_i$:s are empty their union is still in $\tau$.
The union you have written is all of $\mathbb{R}$ which is in the topology.