I have to show the following:
Let $ G=(V,R,\alpha,\omega) $ be a weakly connected, directed graph with more than one node. There exists a node $ v\in V $, such that $ G-v $ ist weakly connected.
We defined weak connectivity as follows:
A directed graph $ G $ is weakly connected, if its underlying graph $ H $ is connected.
I have no idea how to prove this, can someone help me?
An undirected graph is connected if and only if there is a walk on the graph that visits each node at least once. (A walk is like a path but it may repeat nodes and edges). Let $W$ be a minimum length walk on $H$ that visits every node. If $v$ is the last (or for that matter the first) node visited by the walk $W$, then $W$ visits $v$ only once, so $W-v$ is a walk on the graph $H-v$ which visits every node of $H-v$, showing that $H-v$ is connected.
Pick two nodes $u$ and $v$ as far away from each other as possible, i.e., the distance $d(u,v)$ is maximized. (It's equal to the so-called diameter of the undirected graph $H$.) Show that $H-u$ and $H-v$ are connected graphs.
A connected graph has a spanning tree. Let $T$ be a spanning tree for $H$. A tree with more than one node has at least two leaves. If $v$ is a leaf of $T$, then $T-v$ and $H-v$ are connected graphs.