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It is well known that a graph is said to be $k$-partite if its vertex set can be partitioned in to k non empty sets such that no two vertices in the same set are adjacent.

My question is whether such a '$k$' must be minimum ? ; but the definition doesn't have such a condition explicitly.

A reason for this question is as follows: If we do not assume this condition, then every graph on $k $ vertices is $k$-partite, with all the singleton subsets of its vertex set as the partitions. Likewise, the totally disconnected graph on $k$ vertices ( i.e the graph on $k$ vertices having no edges) can be viewed as a r- partite graph for each $1 \leq r \leq k$.

2 Answers 2

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It is less restrictive the larger $k$ gets, so less interesting, but it is true. Given a bipartite graph, you can arbitrarily split one of the parts in two and make a $3-$partite graph. If you prove a theorem about $3-$partite graphs it will apply to bipartite graphs (at least those with at least $3$ vertices).

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    In order to prove a graph to be $k$-partite, is it enough to find a partition of the vertex set having k subsets satisfying the condition that no two vertices in the same sets are adjacent ? – 2017-01-02
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    Yes, that is the definition. I then suggested how, if you have a $k-$ partite graph you could prove it $m-$ partite for any $m \gt k$ until you run out of vertices. – 2017-01-02
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No it does not, for example, every bipartite graph is also $5$-partite.

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    @ Jorge: But a bipartite graph is 2-partite. – 2017-01-02
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    exactly, if it is $k$-partiite it is also $s-$partite for all $s>k$. – 2017-01-02