Let $T$ be a tree with $3$ edges. Let $G$ be a simple graph such that each vertex has degree at least $3$. Show that $G$ has $T$ as a subgraph.
This statement is obvious but I am not sure how to prove it rigorously.
Could anybody please help me check whether my proof is good enough or not, and some advice for improvement if possible. I really think my proof is not good enough, because I am not sure how to fill in the details to make the proof more convincing. Thanks!
Since $T$ is a tree with $3$ edges, then each vertex of $T$ has at least $1$ edge and at most $3$ edges. Then we can extend $T$ by adding edges and vertices so that it becomes $G$, it possible because $G$ has degree at least $3$.
EDITED
We can find a vertex in $T$ that has degree less than $3$, then we can connect that vertex with an edge to another vertex that has degree less than $3$. But we have to make sure there is no loop created. We can keep adding edges so that all vertices have at least $3$ edges. But $G$ is given, so we have to add all the edges according to $G$.
My other concerns are (out of curiosity): as the number of edges increases to a general $n$ edges, then we need to deal with each case of possible graphs with $n$ edges, is there a better way besides dealing with each possible shape of the tree?
Let's say $T$ has 5 edges, then there are more than two trees that has 5 edges, does that mean we have to deal with each case and extend the tree from each case? Is there any better way?
You're right, that's not very convincing. :-)
There are essentially two different trees with $3$ edges. You can have all three edges incident at the same vertex; this is the star graph $S_3$. Or you can have at most $2$ edges incident at any vertex – then the tree is the path graph $P_4$ (why?).
$S_3$ is easy. The claim isn't quite right since it doesn't hold for the empty graph; but if we assume that $G$ is not empty, then it has at least one vertex of degree at least $3$, and any three edges incident at that vertex induce a subgraph isomorphic to $S_3$.
$P_4$ requires a bit more work – I'll leave that to you...