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Prove that $K_{3,3}$ is not planar.

This problem comes from A Course in Combinatorics (Problem 1D). At this point, the authors have introduced merely the most basic concepts of graphs, so this is preferably solved without other results. The hint says: "Call the vertices $a_1,a_2,a_3$ and $b_1,b_2,b_3$. First, omit $a_3$ and show that there is only one way to draw the graph with the remaining $6$ edges in the plane." But how am I supposed to show that there is only one way to draw the graph, since even the locations of the points themselves are arbitrary?

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The edges cannot cross, so if you take $a_1, a_2, b_1, b_2$ you get a quadrilateral. The only choice that you have is that you can put $a_3$ and $b_3$ either outside or inside, which gives you at most 4 cases (actually inside and outside are symmetric so you can reduce the number of cases, but you don't have to use that).

Another way to look at this is: take any planar embedding of your graph, and take a subgraph on $a_1,a_2,b_1,b_2$. It will give you a quadrilateral, so let this quadrilateral of yours (from the previous paragraph) represent the one from the embedding. Now, the embedding could have put $a_3$ and $b_3$ either inside or outside, and so on...

Finally you will get that wherever you would like to put in $b_3$ (one case for each face of the graph on $a_1,a_2,a_3,b_1,b_2$), you cannot draw all the edges without crossing, thus the initial embedding could not exist.

I hope this helps $\ddot\smile$

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  • $\begingroup$ I suppose in order to make the proof completely rigorous one would need to appeal to the Jordan curve theorem at several points. $\endgroup$ – Henning Makholm Feb 9 '17 at 10:39
  • $\begingroup$ @HenningMakholm I don't think you need the (whole) Jordan curve theorem to prove it. That theorem states more than you probably need to prove this. $\endgroup$ – skyking Feb 9 '17 at 10:50
  • $\begingroup$ @HenningMakholm I think that is the easiest way. One alternative would be to prove that if there is an embedding, then there is an embedding that uses only integral points, and create some weaker form of Jordan curve theorem where you can assume your space is finite. $\endgroup$ – dtldarek Feb 9 '17 at 10:51
  • $\begingroup$ Actually this is the kind of answer I was looking for, but how can I make it rigorous? I haven't learned about the Jordan curve theorem before. Is it just that the hint does not lead to an "easy" (in the sense that it does not require deep results from analysis, topology or geometry) proof? $\endgroup$ – Colescu Feb 9 '17 at 13:32
  • $\begingroup$ @YuxiaoXie If the context is topology, then the Jordan curve theorem or some substitute is actually what you should use. However, if the context is graph theory, that part is usually dismissed as "obvious" or "not part of the course". Observe that people are using numbers everyday, but do not feel compelled to prove their properties from axioms every time – that part belongs somewhere else. $\endgroup$ – dtldarek Feb 9 '17 at 15:53
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If you have proved the Euler characteristic of the plane, this can be used for a slicker proof.

Euler's equation $V-E+F=2$ implies that a planar drawing of a graph with $6$ vertices and $9$ edges must have exactly $5$ faces (including the infinite "outside" face).

$K_{3,3}$ is simple, so no face has two sides, and bipartite, so every face has an even number of sides. Thus every face must have at least $4$ sides. But the sum of sides over all the faces is twice the number of edges (each edge is a side of exactly two faces), so this says that $2E\ge 5\cdot 4=20$. But actually $2E$ is $18$, a contradiction.

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  • $\begingroup$ I know this method, but I'm just curious in how the hint leads to a (rigorous) solution. Thank you for your proof anyway! $\endgroup$ – Colescu Feb 9 '17 at 13:28

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