Exlamation about a claim of an existing such cycle in a simple Graph

Suppose the following situation: this is found at (Let $G$ be a graph of minimum degree $k>1$. Show that $G$ has a cycle of length at least $k+1$)

Let $P=v_0v_1 \dots v_l$ be a longest path in $G$. $v_0$ has to have additional neighbors by the degree constraint. All of the neighbors of $v_0$ have to be in $P$, otherwise $P$ could be extended. Therefore $v_0$ has at least $k$ neighbors in $P$. Let $j$ be the maximum index of a neighbor of $v_0$. By the previous statement we have that $j \ge k$. Thus we have the cycle $v_0v_1 \dots v_jv_0$, which has length at least $k+1$.

After many interpretation I did not understand what he meant by $j$ is the maximum index of a neighbor. I assumed it to be directed graph ? In addition I could not understand why $j\geq k$ and why has length at least k+1. I tried but I could not. ?

We know that $v_0$ has at least $k$ neighbors in $P$. Now these neighbors have indices drawn from $\{1, \dots, l\}$ by definition. So by maximum index, he means let $j$ be the index such that $v_j$ is a neighbor of $v_0$ and $v_i$ is not a neighbor of $v_0$ for all $i > j$.
Now $j \ge k$ since if not, then $v_0$ does not have $k$ neighbors in $P$ which is a contradiction. You should be able to figure out the length argument from here.
• if $v_0$ is connected with $v_x$ and also with $v_y$ what is the rule to choose a maximum index is it $x$ or $y$ ? there must be some rules to classify $j$ as maximum index ? what is the idea of choosing indexes here ? – John Kanti Jun 24 '15 at 3:34
• @JohnKanti I'm not sure I follow. Do you mean assume that $v_0$ has neighbors $v_x$ and $v_y$, how do you know what the maximum index is? Well that depends on $x$ and $y$. If $x > y$ then $x$ is the maximum index. If $y > x$, the $y$ is the maximum index. – Ebearr Jun 24 '15 at 3:39
• as I understand you numbered every node, because both of $x$ and $y$ is the label of the node, they are not numbers see please this picture [(i.stack.imgur.com/aB6cH.jpg)] – Tandee Holwa Jun 24 '15 at 3:55
• @TandeeHolwa The vertices in this problem are not like what is in your picture. The vertices in $P$ are $v_0, v_1, \dots, v_l$. – Ebearr Jun 24 '15 at 3:59