Inductively defined sequence of graph neighborhoods I'm trying to solve the following problem:

Let $v_0$ be a vertex in a graph $G$, and $D_0 := \{v_0\}$.
  
  
*
  
*For $n = 1, 2 \dots$ inductively define
  $D_n := N(D_0 \cup D_1 \cup \dots \cup D_{n-1})$.
  
*Show that $D_n = \{v | d(v_0, v) = n\}$ and
  $D_{n+1} \subseteq N(D_n) \subseteq D_{n-1} \cup D_{n+1}$ for all
  $n \in \mathbb{N}$.

Here, $N_G(V') = N(V')$ is the neighborhood of $V' \subseteq V(G)$, and
$d_G(u, v) = d(u, v)$ is the distance in $G$ of two vertices $u$, $v$.
I'm unable to solve the task because I am not sure if it's correctly defined. Every type of graph seems to be a contradiction to the statements above. Consider for simplicity the following example:
Let $P$ be a path, and $v_0$ be the central vertex in $P$. Then
$D_0 = \{v_0\}$,
$D_1 = N(D_0) = \{v | d(v_0, v) = 1\}$,
$D_2 = N(D_0 \cup D_1) = N(\{v_0\} \cup N(\{v_0\})) =
\{v | d(v_0, v) \leq 2\}$,
and for all $n \geq 2$ we obtain $D_n = \{v | d(v_0, v) \leq n\}$. In particular, it follows that $V(P) = D_r$, where $r$ is the radius of $P$.
This example contradicts both statements in the second part of the task. Can you, please, help me find out, is there anything I am missing or interpreting incorrectly?
 A: You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0\cup\ldots\cup D_{n-1})$, they are implying the open neighbourhood
The open neighbourhood $N(S)$ of some set of vertices $S$ does not include $S$. Therefore, using your exampla of the path $P$:
$D_0=\{v_0\}$
$D_1=N(D_0)=\{v \mid d(v_0,v)=1\}$
$D_2=N(D_0\cup D_1)=\{v \mid d(v_0,v)=2\}$
and so on. 
More precisely if you define $P$ as:
$$ \ldots \ v_{-n} \ldots \ v_{-2} \ v_1 \ v_0 \ v_1 \ v_2 \ldots \ v_n \ldots$$
Then $D_0=\{v_0\}$, $D_1=N(D_0)=\{v_1,v_{-1}\}$, and
$$D_2=N(D_0\cup D_1)= N(\{v_1,v_0,v_{-1}\})=\{v_2,v_{-2}\}$$ 
Therefore the statements holds as
\begin{align*}
D_{n+1}=\{v_{n+1},v_{-(n+1)}\} &\subset N(D_n)=N(\{v_{n},v_{-n}\})=\{v_{n+1},v_{n-1},v_{-(n-1)},v_{-(n+1)}\}\\
&\subset D_{n-1} \cup D_{n+1}
\end{align*}
with an equality in your specific case of $P$
A: It depends on how you define the neighbourhood, but it seems that here it is defined so that $N(V')$ comprises the vertices adjacent to some vertex in $V'$, excluding vertices in $V'$.
So in that case, $N(D_0\cup D_1)\neq \{v|d(v_0,v)\leq 2\}$.
