Nikiel's conjecture

About 40 years old · traced to

Let XX be a compact Hausdorff space.

Call a linearly ordered set (L,<)(L,<) with its open-interval topology a linearly ordered topological space, and say that XX is a continuous image of a compact linearly ordered space if there exist a compact linearly ordered topological space LL and a continuous surjection f ⁣:L→Xf\colon L \to X.

Call a T1T_1 space XX monotonically normal if there is an operator HH assigning to each pair (p,U)(p,U), where U⊆XU \subseteq X is open and p∈Up \in U, an open set H(p,U)H(p,U) such that

  1. p∈H(p,U)⊆Up \in H(p,U) \subseteq U; 2. if U⊆VU \subseteq V are open and p∈Up \in U, then H(p,U)⊆H(p,V)H(p,U) \subseteq H(p,V); 3. if p≠qp \neq q and U,VU, V are open with p∈Up \in U, q∈Vq \in V, then
H(p,U)∩H(q,V)≠∅  ⟹  p∈V or q∈U.H(p,U) \cap H(q,V) \neq \emptyset \implies p \in V \text{ or } q \in U.

Then XX is a continuous image of a compact linearly ordered space if and only if XX is monotonically normal.

References

Primary source

Wikipedia

Additional references

  1. Wikipedia, Nikiel's conjecture, the article this problem comes from.

Progress summary

Refreshed
Claimed solved

Later mathematical literature says Mary Ellen Rudin solved Nikiel’s conjecture in 2001, although this scan does not independently verify the proof.

Nikiel’s conjecture asserts an equivalence between a compact space being monotonically normal and being a continuous image of a compact ordered space.

Rudin’s 2001 solution

Several later papers state that Mary Ellen Rudin proved the conjecture in 2001, in its full compact Hausdorff form; one also uses the theorem to derive a connected version. The scan found no competing proof, counterexample, reported gap, or retraction.

Current status (as of August 2026): The conjecture is reported as solved by Mary Ellen Rudin in 2001, but the proof is unverified in this automated record.

Sources

Solutions 0

No solutions have been posted yet.