Let $G = (V, E)$ be an undirected graph, then if $G$ is acyclic, then $|E| ≤ |V | − 1$ Let $G = (V, E)$ be an undirected graph. I want to show that if
 $G$ is acyclic, then $|E| ≤ |V | − 1$.
Attempt:
Suppose by contradiction that $|E| > |V| - 1 \rightarrow |E| \geq |V|$, then the graph of $G$ must contain multiple edges (i.e. for example for vertices $v_1,v_2$ there would be two edges from $v_1 \rightarrow v_2$) or there would be a cycle. 
I want to make this last statement "there would be a cycle" more rigorous. Any hints or insights appreciated.
Also as a side note I find it confusing that in an undirected graph that we could say anything is acylic since we could consider going from one vertex to the next, and then going back, making a cycle? I guess this is not allowed.
 A: Suppose to the contrary, and let $G$ be such a graph of minimal number of edges. Remove an edge, $e$, and you are left with $n\geq 2$ connected components (otherwise, if $e$'s vertices are $v_1$ and $v_2$, trace a path, $p$ from $v_1$ to $v_2$ in the reduced graph, and then add $e$ to it to arrive at a cycle in the original graph), each of which satisfies $|E_i|\leq|V_i|-1$. Then
$$
|V| = |V_1|+\cdots+|V_n|\geq |E_i|+\cdots|E_n| + n = |E|-1+n > |E|
$$
A: Let $G=(V,E)$ be an undirected finite graph with at least as many edges as vertices. We have to show that $G$ contains a cycle.
Observe that, if $v$ is a vertex of degree $0$ or $1$ in $G,$ then deleting the vertex $v$ (along with any edge incident with $v$) results in a subgraph $G-v$ which still has at least as many edges as vertices. Repeating this process, we eventually reach a subgraph $H$ in which every vertex has degree at least $2.$
A graph in which every vertex has degree at least $2$ contains a cycle. Pick any vertex as your starting point, and start walking along the edges of the graph, without repeating edges. Every time you enter a vertex for the first time, you will be able to leave it by a different edge, because the degree is at least $2.$ Since the graph is finite, you will eventually come to a vertex you've seen before, thus making a cycle.
