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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.

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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?

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  • $\begingroup$ 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
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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:

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  • $\begingroup$ 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

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