Status
On this page
Status
Topics
Status
On this page
Status
Topics
Let be a graph on vertices which does not contain a trivial (empty or complete) graph on more than vertices. Must contain at least many induced subgraphs which are not pairwise isomorphic?
Source: erdosproblems.com/1036
An accepted solution exists. The statement is true.
Proved. The site credits Shelah [Sh98], whose Theorem 1.3 proves
the conjecture of Erdős and Rényi in the question's form: for every
there is such that, for large, a graph on vertices with
no complete or edgeless subgraph on vertices has at least
induced subgraphs up to isomorphism (J. Combin. Theory Ser. A 82 (1998),
179--185; refereed; arXiv:math/9707226). The claim page
Shelah records
it as accepted on the refereed venue and the site's acceptance after the forum
comment of 13 September 2025; the site's Lean suffix is its catalog label for
the external Lean development that declares itself a formalization of Shelah's
theorem, carried as a formalization link on that claim page and described
under Formalization below, not built or audited here. The proof-claim tab is
empty, and nothing is independently reviewed here. The site also records Alon
and Hajnal's bound [AlHa91], which falls
short of for every graph the question concerns and so settles
no case of it, and the Erdős--Hajnal theorem [ErHa89b], which answers the
question yes for the graphs in which neither nor its complement contains
, a special case recorded on its own partial claim page,
Erdős and Hajnal.