Show for all continuous $f: (\mathbb{N}, \tau_{c}) \to (\mathbb{R}, \tau_{ST})$, $f$ is a constant function Show for all continuous  , $f$ is a constant function.
where $\tau_{c} = \{A \subseteq \mathbb{N} \mid A = \emptyset \vee \mathbb{N}\backslash A: finite  \}$ the cofinite topology on the natural numbers.
and $\tau_{ST}\,\, \text{is defined as} \,\,(a,b) := \{ x \in \mathbb{R} \mid a < x < b  \}$ the standard topology on the reals.
WTS: that all continuous functions $f: (\mathbb{N}, \tau_{c}) \to (\mathbb{R}, \tau_{ST})$ are the constant functions 
Let $f$ be a continuous function.
So then by assumption $f^{-1} (U) $ is open in $\mathbb{N}$ for all open $U \subseteq \mathbb{R}$
Could use a hint. I was trying to use an open interval in the reals but didn't work for me. 
 A: Hint: Notice that every open set of $\Bbb{N}$ is either empty or the complement of an infinite set. If $f$ is continuous, what does this say about the preimage of any interval?
Solution:

 Let $f$ be continuous. Every interval in $\Bbb{R}$ with nonempty preimage contains all but finitely many elements of $\Bbb{N}$. Therefore, if two intervals are disjoint, then at least one of them must have empty preimage. If the image of $f$ has more than one element in its image, then it cannot be continuous, as any two points in $\Bbb{R}$ can be separated by disjoint open intervals.

A: Suppose $f$ is continuous and non-constant. Then there are $y\neq z$ in the image of $f$. 
Since $\mathbb R$ is Hausdorff in the standard topology we may choose disjoint open sets $U,V$ s.t. $y\in U$and $z\in V$. 
Now continuity of $f$ implies there is an $N\in \mathbb N$ s.t. $n>N \Rightarrow \left \{ n,n+1,n+2\cdots , \right \}\subseteq f^{-1}(U)\cap f^{-1}(V)$, which is impossible since $U$ and $V$ are disjoint.
A: Let $f$ be continuous but assume it is not constant then there exists $x , y \in \mathbb{R}$ s.t. we can find disjoint non-empty open balls (intervals) in $\mathbb{R}$, say $A, B$ such that $x \in A$ and $y \in B$. By assumption of continuity the preimage of each is open i.e. $f^{-1}(A)$ & $f^{-1}(B)$ are open in $\mathbb{N}$ further they will also be disjoint and non-empty by the fact $x$ and $y$ have preimages. So they are cofinite but they are disjoint so they have an empty intersection. $\rightarrow \leftarrow$ $\emptyset = f^{-1} (A \cap B) = f^{-1}(A) \cap f^{-1}(B), f^{-1}(A) \&f^{-1}(B)$ are co-finite so their complements are finite but the complement of the empty set is not finite and they themselves are non-empty thus contradiction.  
