Question

Let k be a fixed positive integer and let G = (V, E) be a loop-free undirected graph, where deg(v)k\operatorname{deg}(\mathrm{v}) \geq \mathrm{k} for all vV.\mathrm{v} \in \mathrm{V}. Prove that G contains a path of length k.

Solution

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Proof by induction.

For k=1k = 1 the claim is trivial.

Suppose that the claim holds for some kNk \in \mathbb{N}.

Let now GG be any graph such that degvk+1\deg v \ge k+1 for every vVv \in V. We then specially have that degvk\deg v \ge k for every vVv \in V, so we can apply the above assumption on GG.

By assumption there exists a path of length kk in GG which passes through the vertices v0v_0, ..., vkv_k. Now since degvkk+1\deg v_k \ge k+1, it follows that there exists a vertex vk+1viv_{k+1} \neq v_i for all iki \le k which is adjacent to vkv_k.

Now the path passing through v0v_0, ..., vk+1v_{k+1} is a path of length k+1k+1 in GG, which proves that the claim holds for k+1k+1.

Now by the induction principle, the claim holds for every kNk \in \mathbb{N}.

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