A non-hereditary Pollyanna class that is not strongly Pollyanna
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arXiv
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| Format: | Preprint |
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2026
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| _version_ | 1866909042259001344 |
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| author | Chen, Hongzhang Lan, Kaiyang |
| author_facet | Chen, Hongzhang Lan, Kaiyang |
| contents | Chudnovsky, Cook, Davies, and Oum introduced the notion of Pollyanna graph classes: a class $\mathcal{C}$ is Pollyanna if for every $χ$-bounded class $\mathcal{F}$, the intersection $\mathcal{C} \cap \mathcal{F}$ is polynomially $χ$-bounded. They further defined $\mathcal{C}$ to be strongly Pollyanna if it is $k$-strongly Pollyanna for some integer $k$, meaning that $\mathcal{C} \cap \mathcal{F}$ is polynomially $χ$-bounded for every $k$-good class $\mathcal{F}$. They asked whether there are Pollyanna graph classes that are not strongly Pollyanna. In this note we answer this question affirmatively, under the literal interpretation that graph classes are not required to be hereditary. We construct a class $\mathcal{C}$ that is Pollyanna but, for every $k \ge 1$, is not $k$-strongly Pollyanna; in particular $\mathcal{C}$ is not strongly Pollyanna. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2605_14547 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | A non-hereditary Pollyanna class that is not strongly Pollyanna Chen, Hongzhang Lan, Kaiyang Combinatorics 05C15, 05C75 Chudnovsky, Cook, Davies, and Oum introduced the notion of Pollyanna graph classes: a class $\mathcal{C}$ is Pollyanna if for every $χ$-bounded class $\mathcal{F}$, the intersection $\mathcal{C} \cap \mathcal{F}$ is polynomially $χ$-bounded. They further defined $\mathcal{C}$ to be strongly Pollyanna if it is $k$-strongly Pollyanna for some integer $k$, meaning that $\mathcal{C} \cap \mathcal{F}$ is polynomially $χ$-bounded for every $k$-good class $\mathcal{F}$. They asked whether there are Pollyanna graph classes that are not strongly Pollyanna. In this note we answer this question affirmatively, under the literal interpretation that graph classes are not required to be hereditary. We construct a class $\mathcal{C}$ that is Pollyanna but, for every $k \ge 1$, is not $k$-strongly Pollyanna; in particular $\mathcal{C}$ is not strongly Pollyanna. |
| title | A non-hereditary Pollyanna class that is not strongly Pollyanna |
| topic | Combinatorics 05C15, 05C75 |
| url | https://arxiv.org/abs/2605.14547 |