Are subgraphs of a minimum spanning tree, also minimum spanning trees? Clarification: MST = minimum spanning tree
Say I take some complete, weighted graph $G$ - I create a MST of the graph then split $G$ into 2 trees $T_1$ and $T_2$ by removing some edge connecting them called $uv$. My question is, that if I were to take the subgraph of all nodes constuting $T_1$ and $T_2$ and the edges originally connecting them in $G$, would the MSTs of these subgraphs be equivalent to the original MSTs $T_1$ and $T_2$? That is, are the subgraphs of a MST equal to the MSTs of the subgraphs?
 A: I will prove the following Lemma, which answers the question you are asking.

$G$ be a graph with vertex set $V$, and let $T$ be a  minimum spanning tree of $V$. Let $T_1$ be any subtree of $T$, and let $W$ be the vertex set of $T_1$. Then $T_1$ is also an MST of $G[W]$, the subgraph of $G$ induced by $W$. 

To prove this, let $T_1'$ be any other spanning tree of $G[W]$. Consider the graph $T'$, with vertex set $V$, whose edge set $E(T')$ is given by
$$
E(T')=(E(T)\setminus E(T_1))\cup E(T_1')
$$ 
That is, take $T$, remove the edges of $T_1$, then add in the edges of $T_1'$. I claim that $T'$ is a spanning tree of $G$. To see this, note


*

*$T'$ is connected. Indeed, for any vertices $v_1,v_2\in V$, there is a path in $T$ connecting them, which can be modified to be a path in $T'$ by replacing the portion of the path that lies in $T_1$ with a corresponding path in $T_1'$. 

*$T'$ has the same number of edges as $T$. 
Since $T'$ is a connected graph with $|V|-1$ edges, it is a tree on the same vertex set as $G$, and therefore a spanning tree. 
Now, since $T$ was an MST for $G$, we conclude that $w(T')\ge w(T)$. Since $w(T)=w(T_1)+w(T\setminus T_1)$, and $w(T')=w(T_1')+w(T\setminus T_1)$, it follows that $w(T_1')\ge w(T_1)$, thus proving the minimality of $T_1$.
A: The answer is yes; if the subgraphs created by taking the nodes and edges from $T_1$ and $T_2$ had a smaller weight than the subgraphs of the minimum spanning tree, and were combined with the original graph by adding the edge $uv$ back, the resulting graph would be a minimum spanning tree of $G$ but with a smaller weight than the original minimum spanning tree created for $G$.
This leads to a contradiction as the definition of a minimum spanning tree is such that the sum of the weights of the edges used to construct it are a minimum, hence, a smaller minimum spanning tree for $G$ cannot exist - therefore, $T_1$ and $T_2$ have to have already been minimum spanning trees as the sum of the weights of the edges in them cannot be decreased and they are already connected trees - hence fitting the definition of a minimum spanning tree.
