Gaining intuition on a statement of ordered minimum spanning tree edges containing no cycles from graph theory

Question

Let G be a connected non-null graph Let $w: E(G) \rightarrow \mathbb{R}_{+}$s.t. $w(e) \neq w(f)$ whenever $e \neq f$. Let $T$ be a minimum spanning tree for $(G, W)$, s.t $E(T)=\left\{e_1, e_2, \ldots, e_k\right\} \quad w\left(e_1\right)<w\left(e_2\right)<\ldots<w\left(e_k\right)$ Then for every $1 \leqslant i \leqslant k$ $e_i$ is the edge of $G$ of minimum weight subject to

• $e_i \notin\left\{e_1, e_2, \ldots, e_{i-1}\right\}$
• $\left\{e_1, e_2, \ldots, e_i\right\}$ does not contain edge set of a cycle.

I am having trouble parsing the meaning of this statement. Does this mean, if you find the minimum spanning tree according to weights, order the edges of it based on their weights, then the first $i$ will be all in a row on $G$ and not form a cycle? This statement is very confusing ot me intuitively and I am unsure when I would even apply it. Moreover, how could there even be a cycle if these are all edges of the spanning tree?

Answer 1

The statement and your attempt on it are both slightly convoluted. The real meaning of the statement is that this algorithm is correct:

Start with an empty edge set $T$. Add $G$'s edges to $T$ one-by-one in order of increasing weight iff the result is acyclic (discard those edges that would form a cycle in $T$ once added). $T$ at the end is a (indeed, the) minimum spanning tree.

This is Kruskal's algorithm.