Topology with Cofiniteness Is the function $f: X \rightarrow Y$ continuous?
X = Y = $\mathbb{R}$ with the topology $\tau$ = { A : $\mathbb{R}$ \ A is finite } $\cup$ {$\emptyset$}, $f(x) = \sin(x)$.
The way it's shown is: 
Let U = $\mathbb{R}$ \ {$0$} then U is open. So $f^{-1}(0) = n\pi $ for $n \in \mathbb{Z}$ and hence $\mathbb{R}$ \ $f^{-1}(0)$ is not open.
I understand the basics of how to show continuity and what a topology is. But not the start of this particular example. I hope you can help me.
1) Doesn't U have to have to fit the condition of the topology or can it be an arbitrary open set? Because $\mathbb{R}$ \ {$0$} is open but NOT finite for my eyes.
2) If U has to fulfill the condition of the topology, how is $\mathbb{R}$ \ {$0$} finite? Has it something to do with the definition "Any intersection of finitely many elements of τ is an element of τ"? Does the topology require finiteness, so even $\mathbb{R}$ \ {$0$} is finite?
3) Why is it not open at the end? I see that $f^{-1}(0)$ is infinite, but it's not the whole space and $\mathbb{R}-f^{-1}(0)$ is still open.
 A: The example is worded somewhat badly so I understand your confusion. 
$U=\mathbb{R} \setminus \{0\}$ is most definitely not a finite set, but its complement $\mathbb{R} \setminus U=\mathbb{R}\setminus (\mathbb{R} \setminus \{0\})=\{0\}$ certainly is. This makes the set $\{0\}$ closed by definition. In other words, $\tau$ is the collection of all subsets of $\mathbb{R}$ which have finite complements, not all subsets which are finite complements.
Then since $f^{-1}(0)=\{n\pi|n \in \mathbb{Z}\}$ is an infinite set, $\mathbb{R} \setminus f^{-1}(0)$ is also infinite (we're taking away only countably many points from $\mathbb{R}$) so its complement is $f^{-1}(0)$, thus it's not included in the topology. Now as for the continuity part, if we take an arbitrary open set $V$, $f^{-1}(V)$ must be open, or dually; if we take an arbitrary closed set $C$ (that is, a finite set or $\mathbb{R}$ itself), $f^{-1}(C)$ must be closed. This fails with $\{0\}$, as $\mathbb{R} \setminus f^{-1}(0)$ isn't open so its complement isn't closed; hence $f(x)$ is not continuous everywhere.
In the end if a set is open with respect to some topology, it must always satisfy the conditions of that topology, if there are any.
I hope this clears things up a tiny bit!
