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Definition of cut:

Let $X$ be a set of vertices of a graph $G$. The cut induced by $X$ is the set of all the edges with an endpoint in $X$ and another in $V_G \backslash X$.

On https://en.wikipedia.org/wiki/Cut_(graph_theory) there's such a figure enter image description here

As far as I understood $∂(X)$ is the set of edges highlighted by red, so $|∂(X)| = 2$ edges, but the leftmost vertex (black) has two edges being endpoints of the same colored-vertices (a.k.a set $X$ vertices), what conflicts with the definition

with an endpoint in $X$ and another in $V_G \backslash X$

Does $X \cup V_G \backslash X = V_G$?

k1r1t0
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  • I'm not sure what your confusion is. Say the set $X$ is the three black vertices. Then the cut induced by $X$ is all edges starting in $X$ and ending in not $X$. If an edge remains entirely in $X$ (like the edges from the leftmost vertex), then it isn't part of the cut. – kipf Apr 08 '25 at 10:34
  • Also, for any sets $A,B$, $(A \cup B) \setminus A = B \setminus A$ and $A \cup (B \setminus A) = A \cup B$ (to answer your last question). – kipf Apr 08 '25 at 10:36
  • @kipf, maybe I didn't make it clear, but as far as I understood $X$ is the set of vertices and all these vertices have edges that are endpoints in $X$ and $V_G \backslash X$, but on the figure from wiki the leftmost vertex has endpoints in $X$ and other endpoints in $X$ as well. Let $X$ be a set of vertices colored black (and are endpoints of red edges) and let $F$ be a set of vertices colored white (and are endpoints of red edges), then $X \cup F \neq V_G$, but the definition says that all vertices that are not in $X$ are in $F$ (the edges have endpoints in $X$ and in $X \backslash V_G$) – k1r1t0 Apr 08 '25 at 10:44

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Not all vertices in $X$ are required to be the endpoint of an edge in the cut.

In the Wikipedia diagram, $X$ is the set of all three black vertices, and $V(G) \setminus X$ is its complement: the set of both white vertices. This is our starting point, before we even look at which vertices the graph has.

Then, we define $\partial(X)$ to be the set of all edges that have an endpoint in $X$ and an endpoint in $V(G) \setminus X$, which is all the edges in red.

It's possible that most vertices in $X$ will not be endpoints of any edges in $\partial(X)$. For example, suppose our graph is a path graph, and we take $X$ to be the first half of the path:

$$\bullet - \bullet - \bullet - \bullet - \bullet \color{red}{{} - {}}{\circ} - {\circ} - {\circ} - {\circ} - {\circ}$$

Then $\partial(X)$ consists of only one edge.

That doesn't mean that those vertices in $X$ don't matter. They exist in $X$ so that the edges out of them don't end up in $\partial(X)$. Back to the Wikipedia diagram: the leftmost vertex is kept in $X$, because if it were in $V(G) \setminus X$ instead, then the two edges out of it would also be red edges in $\partial(X)$.

Misha Lavrov
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  • Thanks, that makes sense, but what's the point in keeping all those vertices in $X$, all of them will have degree 0 (those that have no edges in $∂(X)$), if we consider $(X, ∂(X))$? It seems natural to consider a bipartite subgraph with $∂(X)$ edges and two sets of vertices – k1r1t0 Apr 08 '25 at 20:26
  • Well, we don't consider $(X, \partial(X))$. In general, I don't think it's particulaly profitable to think of a cut as a subgraph; we either think of the set of edges $\partial(X)$, or the partition $(X, V(G) \setminus X)$. The first of these is more intuitively the thing we want, but the second is easier to characterize. – Misha Lavrov Apr 08 '25 at 20:43