# Cardinality of fractal without CH? [closed]

This appears unanswered on reddit.

What are possible choices for the cardinality of plane or line fractal without assuming CH?

Are there fractals for which the answer is not easy? (Sieprinski triangle is open on reddit.)

## closed as off-topic by Gerald Edgar, Andrés E. Caicedo, Chris Godsil, Nik Weaver, Gro-TsenFeb 27 '18 at 15:36

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• Cantor proved that closed subsets of Euclidean space obey CH, no need to assume it. But this question does not belong here. Instead it belongs in math.stackexchange.com Note that reddit thread is from 8 years ago – Gerald Edgar Feb 27 '18 at 11:30
• To answer this question, we need to know the precise definition of a fractal. That given by Mandelbrot (fractal dimension, self-similarity and obtained by iterative procedures) is not as precise as necessary. – Taras Banakh Feb 27 '18 at 11:43
• Any closed subset of Euclidean space is either countable, or has the same cardinality as $\mathbb R$. Reference, G. Cantor, Acta Mathematica 4 (1884) 381--392. Nowadays you should find it in any "real analysis" course, perhaps as the generalization known as the Cantor-Bendixson theorem. en.wikipedia.org/wiki/Perfect_set_property – Gerald Edgar Feb 27 '18 at 12:31
• @EvgenyKuznetsov And it can be further extended to analytic sets: if $A$ is an analytic set, then either $A$ is countable or $A$ contains a perfect set (and in particular has size continuum). These results are part of descriptive set theory, which you might be interested in. – Noah Schweber Feb 27 '18 at 13:01
• Why should we insist that fractals are closed or even analytic? To my way of thinking, the essence of being a fractal is self-similarity by a non-isometry. And in this case, one can easily make lots of fractals of intermediate cardinality when CH fails. – Joel David Hamkins Feb 27 '18 at 13:43

Let's define that a subset $F$ of the plane (or other suitable space) is a fractal, if there is a nontrivial group $G$ of homeomorphisms of the space, at least some of which are not isometries, such that $F=\sigma F$ for every $\sigma\in G$. This expresses the self-similarity feature of fractals.

With this definition, it is easy to find fractals $F$ of any desired infinite cardinality less than or equal to the continuum. Indeed, for any group $G$, there will be a fractal $F$ with respect to $G$ of any size between $\max(|G|,\aleph_0)$ and $\frak{c}$, inclusive.

To build $F$, just pick any collection of starting points in the plane, of the right cardinality, and then simply close the set under all the operations of $G$. This closure will have the desired cardinality, and it follows that $F=\sigma F$ for every $\sigma\in G$, and so $F$ is a fractal.

In particular, the procedure produces countable fractals $F$ with respect to any fixed countable group $G$. One simply starts with a single point $\{a\}$ and then closes under the $G$ action.

One can get versions of the Sierpinski triangle this way, by using the group $G$ that describes the fractal symmetry of that triangle.

I might add that this is a common way of drawing fractals on the computer. One just picks a point and then draws all its images, and then their images and so on, under the desired symmetries. Often, one doesn't want to use a whole group of homeomorphisms, but a weaker collection of maps that exhibit the desired self-similarities, such as the contractive maps only, but the same idea works for this case. For example, with the fern above and the Sierpinski triangle, one doesn't usually apply the expansive maps, and so the resulting image is bounded. But even in those cases, one could apply the inverse maps that produce an unbounded set exhibiting the whole group of symmetries. In this way, one gets a more expansive version of the fern or the Sierpinski triangle that continues outward, exhibiting the desired similarities more fully.

• Concerning unbounded fractals for expanding maps look at the appendix of this paper (arxiv.org/pdf/1304.7529.pdf) for a gallery of macro-fractals (in various projections). By the way, all of them are countable. – Taras Banakh Feb 27 '18 at 14:11
• Unfortunately, this notion of a fractal doesn't really match the usual understanding of what a fractal is - for example, any subset of the place with either contains or is disjoint from an open disc will match your definition - take any homeomorphism which is constant outside this disc but nontrivial inside and let $G$ be the group it generates. It's still better than no definition, I suppose. – Wojowu Feb 27 '18 at 14:41
• Well, usually we are looking for a fractal with respect to a given group $G$, such as the ones giving rise to the Sierpinski triangle or the fern. Indeed, for many common fractals, including those two, the maps are affine. So one excludes your kind of example by insisting on a good group. – Joel David Hamkins Feb 27 '18 at 14:50
• For example, one might insist not merely that $G$ has a non-isometry, but that there is some $\sigma$ in $G$ that acts non-isometrically on $F$ and on the complement of $F$. – Joel David Hamkins Feb 27 '18 at 15:43
• Thanks. Does this transfer properties of fractals to groups? Does "fractal dimension of group" makes sense? – joro Feb 27 '18 at 16:26