Is an arbitrary edge cut an cut set? A doubt that arose when I was browsing the following post.

*

*do-cut-set-and-edge-cut-mean-the-same-thing

A cut set of a graph $G$ induced by a partition of $G$'s vertices
into sets $X$ and $Y$ is the set of all edges with one endpoint in $X$
and another endpoint in $Y$.


An edge cut of a connected graph $G$ is a set $S$ of $G$'s edges
such that $G$-$S$ is disconected.


A minimal edge cut of a connected graph $G$ is a set $S$ of $G$'s edges
such that $G$-$S$ is disconected and $G$-$S$' is connected for any
proper subset $S$' of $S$.

Note that in the original post, the edge cut is represented to the minimal edge cut.
From the responses of Harald Hanche-Olsen, it appears that the two concepts (cut set and minimal edge cut) are not equivalent. Edge cuts are only meaningful for connected graphs, but cut set can also be defined for non-connected graphs.
However, for connected graphs, I feel that any cut set is an edge cut(must minimal?). But what about the reverse? Does it also hold true? Can we find a counter-example? That is to say:

*

*For a connected graph $G$, is an edge cut a cut set in $G$?

 A: First, I want to mention that these two definitions are not consistently used. We definitely want to distinguish between the notions of "a set of edges whose removal disconnects the graph", "a minimal set of that kind", and "the set of all edges between a subset of the vertices and its complement", and one of them is usually called an "edge cut", but it varies which one.
For $S,T \subseteq V(G)$, let $[S,T]$ denote the set of all edges in $G$ with one endpoint in $S$ and one endpoint in $T$. Any set $[S,\overline S]$ where neither $S$ nor $\overline S$ is empty is a "cut set" by the definition in the question. We can say the following:

*

*If $\varnothing \subsetneq S \subsetneq V(G)$, then $G - [S, \overline S]$ is not connected: you can no longer get from a vertex in $S$ to a vertex in $\overline S$.

*It is not always true that $[S, \overline S]$ is a minimal set with this property. Consider a path graph going through vertices $v_1, v_2, v_3, v_4$ in that order. Then $[\{v_1, v_3\}, \{v_2, v_4\}]$ contains all three edges, even though only one of them needs to be removed to disconnect the graph.

*If $F$ is an edge set such that $G$ is connected but $G - F$ is not, then $F$ contains a set of the form $[S, \overline S]$ where $\varnothing \subsetneq S \subsetneq V(G)$. Let $S$ be the set of vertices in one connected component of $G-F$. Then $F$ must contain all of $[S, \overline S]$, because $G-F$ cannot have any of those edges. (Accordingly, if $F$ is a minimal set with the property that $G-F$ is not connected, then $F = [S,\overline S]$.)

When $G$ does not start out connected, we can carry on with the same definitions, but then the only minimal edge set $F$ such that $G-F$ is not connected is empty. This is still equal to $[S, \overline S]$ for some nontrivial choices of $S$.
We can also ask about edge sets $F$ such that $G-F$ has more components than $G$. These must also contain an edge set of the form $[S, \overline S]$ - but the reverse implication no longer holds. Take $S$ to be the set of vertices in a connected component of $G$; then $[S, \overline S]$ is empty, so $G - [S,\overline S]$ has the same number of connected components. To get more meaningful answers, it's probably better to work with each connected component of $G$ separately.
