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The set of minimum edge distances to bipartite for subgraphs of size n is bounded above.
A graph on n vertices has at most n choose 2 edges, and deleting all of them
makes the graph bipartite, providing a straightforward upper bound.
For a given graph $G$ and size $n$, this defines the smallest number $k$
such that any subgraph of $G$ on $n$ vertices can be made bipartite by deleting
at most $k$ edges.
This value is optimal because it is the maximum of minEdgeDistToBipartite taken
over all $n$-vertex subgraphs. This means there exists at least one $n$-vertex
subgraph that requires exactly this many edge deletions.
This is Definition 3.1 in [EHS82].
[EHS82] Erdős, P. and Hajnal, A. and Szemerédi, E.,
On almost bipartite large chromatic graphs Theory and practice of combinatorics (1982), 117-123.
Let $f(n)\to \infty$ possibly very slowly.
Is there a graph of infinite chromatic number such that every finite subgraph on $n$
vertices can be made bipartite by deleting at most $f(n)$ edges?
Is there a graph of infinite chromatic number such that every finite subgraph on $n$
vertices can be made bipartite by deleting at most $\sqrt{n}$ edges?
@[categoryresearchopen,AMS5]theoremerdos_74.variants.sqrt:answer(sorry)↔∃(V:Typeu)(G:SimpleGraphV),G.chromaticNumber=⊤∧∀n,G.maxSubgraphEdgeDistToBipartiten≤(n:ℝ).sqrt:=by⊢ True↔∃VG,χ(G)=⊤∧∀(n:ℕ),↑(SimpleGraph.maxSubgraphEdgeDistToBipartiteGn)≤√↑nsorryAll goals completed! 🐙-- TODO(firsching): add the remaining statements/commentsendErdos74