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Consider the Euclidean plane $\mathbb{R}^2$. A figure is a subset of the Euclidean plane. Two figures $S$ and $T$ are said to have the same shape iff there is a composition of an isometry and a dilation on $\mathbb{R}^2$ such that the image of $S$ is $T$. It is easy to prove that "having the same shape" is an equivalence relation on the set of figures. I define a shape to be an equivalence class of that equivalence relation. My question is, what is the cardinality of the set of shapes? I conjecture that it is $2^{2^{\aleph_0}}$. Is this true, and if so, what is the proof?

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    $\begingroup$ Just checking that those really are the conditions you require a "figure" to have - no conditions like "is a closed curve" or anything that would make it look like, well, a shape? $\endgroup$ May 17 at 20:06

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Despite the geometric flavor, this is really a cardinal arithmetic question. Each equivalence class has size at most continuum (since that's how many isometries and dilations there are), and whenever we partition a set of size $2^{2^{\aleph_0}}$ (such as $\mathcal{P}(\mathbb{R}^2)$) into size-$\le 2^{\aleph_0}$ classes we always wind up with $2^{2^{\aleph_0}}$-many classes.

Note that this is very coarse - the result wouldn't change if, for example, we allowed arbitrary continuous self-bijections of $\mathbb{R}^2$ in the definition of shape.

On the other hand, things get (a bit) more interesting if you restrict attention to "nice" shapes. For example, if we only look at closed sets in the plane, then we're partitioning a set of size $2^{\aleph_0}$ into equivalence classes each of which has size $\le 2^{\aleph_0}$, and at that point any number of classes between $1$ and $2^{\aleph_0}$ inclusive is "cardinal-theoretically" possible; we have to actually think about the geometry to get that there are still as many shapes as there could be.

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