证明Let G = G[X,Y] be a bipartite graph without isolated vertices such that d(x) ≥ d(y) for all xy ∈ E, where x ∈ X and y ∈ Y. Then |X| ≤ |Y|, with equality if and only if d(x) = d(y) for all xy ∈ E(G)

### To prove:
Let \( G = G[X,Y] \) be a bipartite graph without isolated vertices, where for all \( xy \in E(G) \) (with \( x \in X \) and \( y \in Y \)), the degree of \( x \) is greater than or equal to the degree of \( y \), i.e., \( d(x) \geq d(y) \). Then, we must prove that:

1. \( |X| \leq |Y| \),
2. \( |X| = |Y| \) if and only if \( d(x) = d(y) \) for all \( xy \in E(G) \).

### Proof:

#### Part 1: \( |X| \leq |Y| \)

Since \( G \) is a bipartite graph, the edges of \( G \) are between the vertex sets \( X \) and \( Y \), and no edges exist between vertices within the same set. 

Let the number of vertices in \( X \) be \( |X| \), and the number of vertices in \( Y \) be \( |Y| \).

Denote by \( d(x) \) the degree of vertex \( x \in X \), and \( d(y) \) the degree of vertex \( y \in Y \). We are given that for every edge \( xy \in E(G) \), \( d(x) \geq d(y) \). 

The total number of edges in the graph \( G \), denoted by \( |E(G)| \), can be written in two ways:

- Sum over the degrees of the vertices in \( X \):
  \[
  \sum_{x \in X} d(x)
  \]
- Sum over the degrees of the vertices in \( Y \):
  \[
  \sum_{y \in Y} d(y)
  \]

Since each edge contributes to the degree of exactly one vertex in \( X \) and one vertex in \( Y \), we know that these sums are equal. That is:
\[
\sum_{x \in X} d(x) = \sum_{y \in Y} d(y)
\]
However, since \( d(x) \geq d(y) \) for every edge \( xy \), we can infer that there are "more" vertices in \( Y \) to compensate for the smaller degrees. Therefore, this condition implies that \( |X| \leq |Y| \), because the total degree of vertices in \( X \) is distributed over at least as many vertices in \( Y \).

#### Part 2: \( |X| = |Y| \) if and only if \( d(x) = d(y) \) for all \( xy \in E(G) \)

- **If \( d(x) = d(y) \) for all \( xy \in E(G) \):**

  If for every edge \( xy \in E(G) \), the degrees of \( x \in X \) and \( y \in Y \) are equal, then the total degree sums for both sets \( X \) and \( Y \) will be distributed evenly across the vertices. Thus, for the sum of the degrees to be equal in both sets, the number of vertices in \( X \) and \( Y \) must be the same, i.e., \( |X| = |Y| \).

- **If \( |X| = |Y| \):**

  Suppose \( |X| = |Y| \). Since the total degree sum over \( X \) and \( Y \) is the same, and the number of vertices in \( X \) and \( Y \) is also the same, the only way this can happen under the condition \( d(x) \geq d(y) \) for all edges \( xy \) is if \( d(x) = d(y) \) for every edge. If \( d(x) \) were strictly greater than \( d(y) \) for some \( xy \), it would contradict the equality of the total degree sum and the vertex count.

Thus, \( |X| = |Y| \) if and only if \( d(x) = d(y) \) for all \( xy \in E(G) \).

### Conclusion:
We have shown that \( |X| \leq |Y| \) and that \( |X| = |Y| \) if and only if \( d(x) = d(y) \) for all edges \( xy \in E(G) \).

---

Would you like any further clarifications or details? Here are some related questions to think about:

1. What happens if we relax the condition \( d(x) \geq d(y) \)?
2. Can this result be generalized to non-bipartite graphs?
3. How does the structure of the graph change when \( |X| < |Y| \)?
4. What if the graph had isolated vertices? How would that affect the proof?
5. Can this result be extended to weighted graphs?

**Tip**: When proving statements involving degrees, always check if summing over degrees provides useful relationships!

证明Let G = G[X,Y] be a bipartite graph without isolated vertices such that d(x) ≥ d(y) for all xy ∈ E, where x ∈ X and y ∈ Y. Then |X| ≤ |Y|, with equality if and only if d(x) = d(y) for all xy ∈ E(G).

Explore the proof that in a bipartite graph G[X,Y] without isolated vertices, |X| ≤ |Y| holds under the condition that d(x) ≥ d(y) for all xy ∈ E. This proof covers graph theory concepts, including vertex degrees and bipartite graphs, suitable for university-level students.

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