Proving equivalent definitions for MSTs I am working on the following homework exercise:

Let $G = (V,E)$ be an undirected graph and $c: E \rightarrow \mathbb{R}$ it's cost function. Further let $T = (V,E')$ be a spanning tree in G.

I need to show the equivalence of the following statements:


$T$ is a MST.
For all $e = \{x,y\} \in E \setminus E'$ holds: No edge of the (unique) path from $x$ to $y$ has a higher cost than $e$.
For all $e \in E'$ holds: Let $C$ one of the two connected components of $T-e$, then $e$ is an edge with minimal weight in $\delta(V(C))$.


I have done $1) \implies 2)$ and $2) \implies 3)$, but I do not know how to prove $3) \implies 1)$. Could you give me a hint?

EDIT:
Please note that $\delta$ refers to the cut, i.e.:

$$\delta(X) = \{ \ \{u,v\} \in E(G) : u \in X \text{ and } v \notin X \}\}$$
In order to make sure $3)\implies1)\,$ is true, let us assume that all costs of edges are different. Otherwise, there is a counterexample at Is a set of acyclic |V| - 1 light edges always a Minimum Spanning Tree?.

Here is a hint to prove $3) \implies 1)\,$. Suppose you have some MST $T'$. Show that $T$ and $T'$ have the same number of edges.  Show that $T'$ must contain every minimal cut.