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Deleting the high-degree vertices of a -self-regular set of density leaves more than of it, and that remainder is -sparse
Statement
Let , let be nonempty, and suppose is a -regular pair of density . Then there is a subset with such that is -sparse.
Facts & Assumptions
Given: A finite simple graph , a real , a nonempty set , and a density such that is -regular.
If is a -regular pair of density and a fixed set has , then fewer than vertices of have more than neighbours in (In a regular pair, fewer than vertices have too small a degree into a large subset, and fewer than have too large a degree, -regular pairs and self-regular vertex sets).
A set is -sparse exactly when every vertex of the induced graph on it has degree at most times its size (A set is -sparse exactly when the maximum degree of the graph it induces is at most times its size, -sparse, -dense and -restricted vertex sets).
Proof
Apply [L1] to the pair with . Since , fewer than vertices of have more than neighbours in .
Let be the remaining vertices. Then .
For every , one has . Therefore [L2] makes -sparse.
Depends on
- $\epsilon$-regular pairs and self-regular vertex sets
- In a regular pair, fewer than $\epsilon|X|$ vertices have too small a degree into a large subset, and fewer than $\epsilon|X|$ have too large a degree
- Edge counts and densities between nonempty vertex sets
- $c$-sparse, $c$-dense and $c$-restricted vertex sets
- A set is $c$-sparse exactly when the maximum degree of the graph it induces is at most $c$ times its size
Used by
Dependency tree · two levels
11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Y. Zhao, Graph Theory and Additive Combinatorics, Lemma 2.2.3 and Remark 2.3.2 (standard reference, not scraped)