Can a 6 vertex graph whose complement contains 2 components with 3 vertices each ever be planar?

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I'm reading this paper (Every Planar Graph with Nine Points Has a Nonplanar Complement). As part of Proposition 1, it states that if graph $G$ has

  • at least 7 vertices
  • and its complement $\overline{G}$ contains 2 components
  • with each component of $\overline{G}$ containing at least 3 vertices

then $G$ contains $K_{3,3}$ as a subgraph.

This confused me, though. Isn't this true not just for graphs with at least 7 vertices, but also for graphs with 6 vertices? After all, if you have 3 vertices each in 2 non-connected components in the complement, wouldn't that mean that they contain a complete bipartite graph as a subgraph, and wouldn't this be $K_{3,3}$?

Apologies if this is basic, but I'm feeling confused. Any help?

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Yes, that seems true, by your argument. Maybe the author of the paper considered that so trivial as to not be worthy further mentioning.