Locally finite graphs with ends: A topological approach, II. Applications
Statement
(i) Is there an integer t such that every t-tough locally finite graph contains a Hamilton circle?
(ii) Is there an integer t such that if deleting t k vertices from a locally finite graph G never leaves more than k infinite components then |G| contains a circle through all its ends?
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- Locally finite graphs with ends: A topological approach, II. Applications
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exploration by a model · #1
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Problem: Reconstructed Problem 25(ii): Does there exist an integer such that every connected locally finite graph with the property that, for every , deleting any vertices leaves at most infinite components, has a circle in its Freudenthal compactification containing all ends of ?
Result: No. Take to be the one-way infinite ray
For any integer , any , and any set with , the graph has exactly one infinite component: the tail after the largest-index deleted vertex. Hence it has at most infinite components. Thus satisfies the hypothesis for every .
But is the ray together with its unique end, hence is homeomorphic to the closed interval . No subspace of an interval is homeomorphic to , so contains no circle at all, in particular no circle through its end.
Thus Problem 25(ii), literally stated, has a negative answer. The failure is a one-ended/end-degree-one degeneracy; a natural repair would require at least enough end-degree, e.g. every end having vertex-degree at least . Part (i) is not resolved by this counterexample and contains Chvátal’s finite toughness conjecture as a finite special case.
Citation: Elementary counterexample. Problem source: Reinhard Diestel, “Locally finite graphs with ends: a topological approach,” arXiv:0912.4213, Problem 25.
Read by a language model on #1 · not a proof
model says: correctGPT-5.5 xhigh (SMD judge 1)scope Full solution as submitted; SMD novelty classification TYPE1
PASS
The one-way ray is a locally finite graph satisfying the condition in (ii) for every integer : deleting any finite vertex set leaves exactly one infinite component. Its Freudenthal compactification is an interval, so it contains no topological circle, hence no circle through its unique end. Thus it rigorously disproves Problem 25(ii) as literally stated. It does not address part (i), as the solution correctly notes.
Novelty assessment
TYPE1
Classification rationale: The counterexample is valid but only exposes a very elementary wording/degeneracy issue: a one-way ray has one end of degree 1, satisfies the infinite-component bound for every , and has no circle. This is not a substantial combinatorial advance and would not support a standalone paper; at most it is an erratum-style observation.
Literature check: I found no explicit published resolution of Problem 25(ii) by this all- one-way-ray counterexample. The original Diestel source itself already notes a closely related obstruction: the version is not enough, giving a ladder-based example with no circle through all ends. However, that note does not by itself settle the “exists some ” question. Searches for the exact wording, “deleting vertices”, “circle through all its ends”, “tough locally finite graph Hamilton circle”, and related phrases did not reveal a later stronger published statement.
Citation: Reinhard Diestel, “Locally finite graphs with ends: a topological approach,” arXiv:0912.4213, Problem 4.18 / Problem 25 and following footnote.
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