Distance vector in a weighted graph I got a weighted, connected and directed graph $G$. There is a vector called the distance vector $Dv \in \mathbb{N}^n$ in which $Dv_i$ is the shortest distance from $1$ to $i$. All edge weights are positive integers. I have to show that every distance vector $Dv$ satisfies:


$Dv_1 = 0$.
For all $j \neq 1$ there exists $i$ such that $Dv_j = Dv_i + w(i,j)$.
For all $i,j$ it holds that $Dv_j \leq Dv_i + w(i,j)$.


I think that 1 is trivial: from 1 to 1 you have no distance. But the rest? Can you give me an idea how to prove 2 and 3?
Hint for 2: If $1,\ldots,i,j$ is a shortest path from $1$ to $j$ then $1,\ldots,i$ is a shortest path from $1$ to $i$.

Hint for 3: If $1,\ldots,i$ is a path from $1$ to $i$ then $1,\ldots,i,j$ is a path from $1$ to $j$.

In both cases $\ldots$ represents some list of vertices.