# Fractal plane continuum with order $\aleph_0$?

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.

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?

• 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. – Gerald Edgar May 1 at 18:16
• @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. – D.S. Lipham May 1 at 18:42

• 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. – D.S. Lipham Jun 2 at 20:01