mjhjmtu

I'm trying to solve the following problem:

Let $v_0$ be a vertex in a graph $G$, and $D_0 := v_0$.

1. For $n = 1, 2 dots$ inductively define
   $D_n := N(D_0 cup D_1 cup dots cup D_n-1)$.




2. Show that $D_n = v $ and
   $D_n+1 subseteq N(D_n) subseteq D_n-1 cup D_n+1$ for all
   $n in mathbbN$.

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

$D_0 = v_0$,

$D_1 = N(D_0) = d(v_0, v) = 1$,

$D_2 = N(D_0 cup D_1) = N(v_0 cup N(v_0)) =
v $,

and for all $n geq 2$ we obtain $D_n = d(v_0, v) leq n$. 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?

You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0cupldotscup D_n-1)$, 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 $P$:

$D_0=v_0$

$D_1=N(D_0)=v mid d(v_0,v)=1$

$D_2=N(D_0cup D_1)=v mid d(v_0,v)=2$
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
$$D_2=N(D_0cup D_1)= N(v_1,v_0,v_-1)=v_2,v_-2$$

Therefore the statements holds as
beginalign*
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
endalign*
with an equality in your specific case of $P$

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, $N(D_0cup D_1)neq d(v_0,v)leq 2$.