How can we show that $\phi$ is constant? Hello everyone I have the following statement that I want to prove but have no idea how I can proceed, if anyone could help it would be a lot appreciated :
Let $\mathbb N = \{1,2,\dots\}$ with the topology where the open sets are $\emptyset$, $\mathbb N$ and $\{1,\dots,n\}$ for all $n \geq 1$. Also, let $\mathbb R$ with the usual topology.

How can one show that a continuous function $\phi : \mathbb N \rightarrow \mathbb R$ is constant?

Here the point is to show that $(\forall n \geq 1) : \phi(n)=k$ but I don't see how can prove it. Thanks in advance for your help.
 A: In $\mathbb R$ with the euclidean topology the singletons $\{a\}$, with $a\in\mathbb R$, are closed sets. By assumption, the function $f:\mathbb N\to\mathbb R$ is continuous, hence the preimage of a closed set is a closed set.
This means that the preimage of a generic singleton $f^{-1}(\{a\})$ is a closed set in $\mathbb N$ equipped with the topology $\mathcal T=\{1,\dots,n\}_{n\ge 1}\cup\mathbb N\cup\emptyset$. 
The complement of open sets are closed sets for definition, so a closed set $K\in(\mathbb N,\mathcal T)$ can be $\mathbb N$, $\emptyset$ or $\mathcal C(\{1,\dots,n\}_{n\ge 1})\implies f^{-1}(\{a\})=\begin{cases}\mathbb N\\\emptyset\\ C(\{1,\dots,n\}_{n\ge 1}) \end{cases}$.
At this point the preimage of a point in $\mathbb R$ must be $\mathbb N$ or $\emptyset$ because in the other case, taking two distinct points $a,b\in\mathbb R$, since the fact they are closed sets, their preimages would be $f^{-1}(a)=\mathbb N$\ $\{1,\dots,n_0\}$ and $f^{-1}(b)=\mathbb N$\ $\{1,\dots,n_1\}$.
Hence $f^{-1}(a)\cap f^{-1}(b)=\mathbb N$\ $\{1,\dots,\max\{n_0,n_1\}\}$, which is a contadiction $\implies f$ constant.
A: Based on the hint of Kavi Rama Murthy, I think this argument might work:
Assume by contradiction that $\phi$ is non-constant. So there exist $a,b \in \Bbb{N}$ such that $\phi(a) \neq \phi(b)$. Since $\Bbb{R}$ is Hausdorff, we can find disjoint open sets $U, V$ such that $\phi(a) \in U, \phi(b) \in V$ and $U \cap V = \emptyset.$
Since $\phi$ is continuous, $\phi^{-1}(U)$ and $\phi^{-1}(V)$ are open in $\Bbb{N}$. Now we use Kavi Rama Murthy's hint: One of these sets must be empty*. Indeed, if both of them were nonempty, say $\phi^{-1}(U) = \{1,2,...,n\}$ and $\phi^{-1}(V) = \{1,2,...,m\}$ for some $ m,n \in \Bbb{N}$ then their intersection would be nonempty. But we have that
$$\phi^{-1}(U) \cap \phi^{-1}(V) = \phi^{-1}(U \cap V) = \emptyset,$$
since the intersection $U \cap V$ is empty by the Hausdorff property.
But that is a problem: If, say, $\phi^{-1}(U)$ is empty then $\phi(a) \notin U$, a contradiction. Therefore $\phi$ must be constant.
* You can already conclude from here (see Henno Brandsma's comment)
