# The subgraph obtained by removing an edge incident to a degree 1 vertex in a connected graph

Suppose that $v$ is a vertex of degree $1$ in a connected graph $G$ and that $e$ is the edge incident on $v$. Let $G′$ be the sub- graph of G obtained by removing $v$ and $e$ from $G$. Must $G′$ be connected? Why?

My solution:
If I understand correctly I think G′ will be disconnected. The reason is that for graph to be connected each pair of vertexes must be connected. By removing one vertex and edge the node will be disconnected.

Is this correct?

• Are you familiar with what it means for $v$ to be degree 1? – dalastboss Jan 4 '15 at 4:11

If the vertex $v$ has degree $1$, then $v$ is incident to only one edge, namely $e$. This means that $v$ is connected to the rest of $G$ by way of only one edge, and $v$ will only have one other vertex connected to it. So removing $v$ and its only incident edge $e$ from connected graph $G$ means what?
• @NadiaS, the thing is that you are removing $e$ (which is a "cut-vertex" or "bridge" by the way). Removing $e$ by itself would disconnect the graph, leaving the rest of $G$ and the last vertex $v$ all by itself. But, in your question, we are also removing $v$. So we disconnect the graph by removing $e$, but then we also remove vertex $v$, which is the only disconnected vertex. Try this: draw a bunch of connected graphs with a vertex $v$ that has degree $1$. Label $v$ and the edge $e$ incident to $v$. Then erase both $v$ and $e$. You will find $G'$ is always connected. – Sultan of Swing Jan 4 '15 at 23:36
Think about the problem a little. What does a vertex of degree $1$ look like? What does a cut-edge look like?