Continuum means compact and connected.

Define the order of a point $x$ in a continuum $X$ to be the least cardinal $\alpha$ such that $X$ has a neighborhood base of open sets at $x$ with no more than $\alpha$ points their boundaries.

enter image description here

The Sierpinski triangle has three points of order $2$, countably many points of order $4$ (the vertices of the other triangles), and all other points are of order $3$.

The Sierpinski carpet has order $\mathfrak c=|\mathbb R|$ at each of its points.

I am looking for something to go between the Sierpinski triangle and the Sierpinski carpet.

Question 1. Is there a fractal plane continuum which has order $\aleph_0$ at each of its points?

Fractal can be loosely interpreted here to mean "self-similar", "simple recursive construction", or "intersection of an easily definable nested sequence of plane domains".

A slight variation on Question 1:

Question 2. Is there a fractal plane continuum which has a basis of open sets with countably infinite boundaries?

  • $\begingroup$ When your title said "order $\omega$" I was expecting some sort of ordinal invariant. If you had said "order $\aleph_0$" I would have expected a cardinal, as you indeed explain. $\endgroup$ – Gerald Edgar May 1 at 18:16
  • $\begingroup$ @GeraldEdgar Thanks for pointing that out. There is also something known as "rim-type" of a continuum which is expressed in ordinals. My property depends on cardinality only. $\endgroup$ – D.S. Lipham May 1 at 18:42

There is an upper semi-continuous decomposition of the Cantor set times the unit interval which should fit the bill for Question 1, but not for Question 2. It lies in the plane and has a simple recursive construction. Here are the first few steps:

enter image description here

  • $\begingroup$ There is a problem at the top and bottom "endpoints" which can be easily resolved. As it stands, these points only have order 1. But the order becomes $\aleph_0$ everywhere if we squeeze the top and bottom Cantor sets to a single point. $\endgroup$ – D.S. Lipham Jun 2 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.