Proving open neighbourhood in topology Let $X$ be the set $(\mathbb{R}\backslash \mathbb{N}) \cup \{1\}$. Define a function $f:\mathbb{R} \rightarrow X$ by 
$$
 f(x) = \left\{ \begin{array}{ll}
         x & \mbox{if $x \in \mathbb{R}\backslash \mathbb{N}$};\\
        1 & \mbox{if $x \in \mathbb{N}$}.\end{array} \right.
$$
Further, define a topology $\tau$ on $X$ by 
$$ \tau = \{U: U\subset X \; \text{and}\; f^{-1}(U) \; \text{is open in the euclidean topology on} \; \mathbb{R} \} $$
I need to prove that Every open neighbourhood of $1$ in $(X,\tau)$ is of the form $(V\backslash \mathbb{N}) \cup \{1\}$, where $V$ is open in $\mathbb{R}$.
My approach was the first showing that $f$ is continuous, and it is obvious by the definition of the topology, $\tau$. Then, I showed that for any open set, $U \in \tau$, ${1} \notin {U}$ because suppose $1 \in U$, and by the definition of topology, $\tau$, $f^{-1}(U)$ need to be open in $\mathbb{R}$. However, since $1\in U$, an inverse mapping of $U$ is $U\cup \mathbb{N}$. Hence, all open sets in $X$ must not contain $1$ as its member. Having said that, by the definition of open neighbourhood of $1$, I need to find an open set in $X$ containing $1$. But since any open set does not contain $1$, so there is no neighbourhood of $1$ in $(X,\tau)$ is of the form $(V\backslash \mathbb{N}) \cup \{1\}$, where $V$ is open in $\mathbb{R}$.
I assume what I have done is wrong as following questions say that $(V\backslash \mathbb{N}) \cup \{1\}$ is open in $X$. Please correct me.
Thanks.
 A: Firstly, I want to highlight that the claim

Hence, all open sets in X must not contain 1 as its member

should immediately raise suspicion. The reason for this is that for $(X,\tau)$ to satisfy the axioms of a topological space, $X$ itself must be an open (and closed) subset of $X$. Since $X$ clearly contains 1, the claim cannot be true. Indeed $f^{-1}(X)=\mathbb{R}$, which is open in the euclidean (any!) topology of $\mathbb{R}$.
As to the question itself, from the definition of $\tau$ we have that if $U$ is an open subset of $X$, $V=f^{-1}(U)$ is open in $\mathbb{R}$. If we further require that $1\in U$ we can consider the fact that the pre-image of every point of $p\in U$ other than $1$ is $p$ itself, while the pre-image of $1$ is $\mathbb{N}$.
This leads us to conclude that $V=U \cup \{2,3,4,...\}$ where the union is disjoint. performing some simple set arithmetic at this point, while recalling that $1\in V$ indeed leads to $(V/\mathbb{N})\cup \{1\} = U$ as desired.
Not that for every neighborhood $U$ there are many sets $V$ in $\mathbb{R}$ that satisfy the desired condition, but we only have to find one such $V$ for every $U$, and I merely thought this method is simplest.
