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In a connected graph $G$ where degree of every vertex $v$ is even, show that $G\setminus v$ has at most $\dfrac{1}{2}\deg(v)$ connected components.

$G\setminus v$ is the graph which is left after removing $v$ and all of its incident edges from $G$.

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The graph $G$ is Eulerian. Fix any Eulerian cycle on $G$. It is easy to see, when we remove $v$ from $G$, the cycle will be cut into at most $\frac 12\deg v$ parts, each of which is connected. Since each connected component of $G\setminus\{v\}$ is a union of these parts, there are at most $\frac 12\deg v$ components.

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