Turán's theorem

How many edges can we retain while excluding a prescribed clique? For integers $r\ge 1$ and $n\ge 0$, every simple graph on $n$ vertices containing no complete subgraph $K_{r+1}$ satisfiesEquality holds exactly for the Turán graph $T_r(n)$, up to graph isomorphism. Thus the extremal number is $\mathrm{ex}(n,K_{r+1})=t_r(n)$.^[ref]

For example, $T_3(8)$ has independent sets of sizes $3,3,2$ and $21$ edges, as counted in Turán graph. It contains triangles in a graph, but no four-vertex clique. Consequently any simple graph on eight vertices with $22$ edges contains $K_4$, while $21$ edges alone do not force one. Excluding larger cliques permits denser graphs; Mantel theorem is precisely the first nontrivial case.

A short proof outline uses Zykov symmetrization. If an extremizer is not a complete multipartite graph, it has vertices $x,y,z$ with $xy,xz$ absent and $yz$ present. If either $y$ or $z$ has larger vertex degree than $x$, clone that vertex in place of $x$. Otherwise replace both $y,z$ by nonadjacent clones of $x$, gaining $2d(x)-d(y)-d(z)+1>0$ edges. Both contradict extremality; balancing the resulting at most $r$ parts gives $T_r(n)$.^[ref]

The vertex cloning in a graph step preserves clique exclusion because any clique using a clone can substitute its original vertex; nonadjacent clones cannot occur together in a clique. At most $r$ parts can be nonempty, since selecting one vertex from each gives a clique. If fewer than $r$ parts are nonempty and one has at least two vertices, splitting it creates more edges, so it cannot be extremal. For $r=1$ the bound is zero. For $r\ge n$, the Turán graph is $K_n$, and the bound is the total $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possible edges.