For every $v\in V$, the MST of $G$, contains an edge, touching $v$, that has the minimal weight from all the edges touching $v$

Question

I'm trying to prove the following statement:

Let $G=\left(V,E\right)$ be an undirect and connected graph and $w\,:\,E\to\mathbb{R}$ weight function on edges. Then for every $v\in V$, the MST of $G$ contains an edge, touching $v$, that has the minimal weight from all the edges touching $v$.

My attempt:

Let $T=\left(V,E_{T}\right)$ be an MST of $G$ by $w$. Let $e=\{u,v\}\in E$ be the edge that has the minimal weight from all the edges touching $v$. Lets assume that $e\not\in E_{T}$. This means that there is an edge $e'=\{w,v\}\in E$ so $e'\in E_{T}$. Since $e$ is the has the minimal weight from all the edges touching $v$, then $w\left(e\right)<w\left(e'\right)$. From that we get $-w\left(e'\right)+w\left(e\right)<0$. Lets take a look at $T'=\left(V,E_{T'}\right)$ so $E_{T'}=\left(E_{T}\backslash\left\{ e'\right\} \right)\cup\left\{ e\right\} $. We get: $$w\left(T'\right)=w\left(T\right)-w\left(e'\right)+w\left(e\right)\overset{-w\left(e'\right)+w\left(e\right)<0}{<}w\left(T\right)+0=w\left(T\right)$$ As a contradiction for $T$ being MST of $G$ by $w$. So we get $e\in E_{T}$.

How is my solution? I feel like I need to explain:

1. Why $w\left(e\right)<w\left(e'\right)$ and not $w\left(e\right)\leq w\left(e'\right)$?
2. Why $T'$ is a tree?

For the first question, if it's true then my whole proof is invalid. Please review my solution and help me answer those questions (or point any other suggestions).

Answer 1

You are almost there.

Ad 1: If $w(e')=w(e)$, then you are actually done as $e'$ is then an edge with minimal weight incident with $v$.

Ad 2: Actually, depending on your choice of $e'$, $T'$ might not always still be a tree. To see this, suppose your graph has vertices $\{1,2,3,4\}$ and edges $\{1,2\}, \{2,3\}, \{1,3\}, \{3,4\}$. Consider the tree made up of the edges $\{1,2\}, \{2,3\}, \{3,4\}$ and suppose in your proof, we have $v=3$ and $e=\{1, 3\}$; as you don't further qualify the choice of $e'$, it might well be that $e'=\{3, 4\}$ - but with this choice of edges, the result $T'$ will not be a tree.

So you have to modify your approach. What happens if you add $e$ to the tree $T$? You will create a unique cycle containing $v$. Now how should you choose $e'$?