# Connected graph definition

It is correct to say that a connected graph is only when there exist some vertex that is connected to all other vertices?

I think this is correct, because a connected graph not all vertices are connected together, but at least no vertex is not connected? Isn't that the above definition is saying?

I know the standard definition but I want to know if this definition is also correct. Thanks

• What do you mean by "vertex $u$ is 'connected' to vertex $v$"? Do you mean that there is a path from $u$ to $v$? Or do you mean that $u$ and $v$ are joined by an edge of the graph?
– bof
Commented Apr 5, 2014 at 21:57
• I mean that in all graphs there is some vertex that is connected to (all) other vertices. Commented Apr 5, 2014 at 22:00
• My question was, what does "connected to" mean?
– bof
Commented Apr 5, 2014 at 22:03
• That there is an edge between two vertices. Commented Apr 5, 2014 at 22:04
• That's what I was afraid you meant. In standard terminology, $u$ is joined (or adjacent) to $v$ if there is an edge $uv$; they are connected if there is a path (possibly containing many edges) from $u$ to $v$. Consider the connected graph with $5$ vertices $u,v,w,x,y$ and $4$ edges $uv,vw,wx,xy$. The vertex $u$ is joined only to $v$, but it's connected to all the other vertices; it's connected to $y$ by the path $u,uv,v,vw,w,wx,x,xy,y$.
– bof
Commented Apr 5, 2014 at 22:11

That is correct. A graph is connected if and only if for all $x, y \in V(G)$, there exists a path from $x$ to $y$.
When talking about directed graphs, we have the concepts of weak connectivity vs. strong connectivity. In a weakly connected graph, we are guaranteed either a directed $x-y$ path or a directed $y-x$ path, but not necessarily both. A strongly connected path guarantees us both a directed $x-y$ path and a directed $y-x$ path.
• This is the definition I'd use. It's a more exacting and stronger definition than the one you gave, no offense: "A graph is connected if and only if for all $x,y \in V(G)$, there exists a path from x to y." Commented Apr 5, 2014 at 21:59