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$\mathcal{P}(\mathbb{R}^2)$ is a set containing as elements all possible collections of points on the plane. What exactly is the power set of this set? $\mathcal{P}(\mathcal{P}(\mathbb{R}^2))$ just contains sets of all possible collections of points on the plane.

In what way does $\mathcal{P}(\mathcal{P}(\mathbb{R}^2))$ differ from $\mathcal{P}(\mathbb{R}^2)$, since the latter by definition already includes every collection of points? What new "content" is the former offering?

I'm aware that the elements of $\mathcal{P}(\mathbb{R}^2)$ are sets of points and the elements of $\mathcal{P}(\mathcal{P}(\mathbb{R}^2))$ are sets of sets of points, so there is a technical difference. But visually on the plane the elements of both sets can be the same?

For example:

$A = \{ (1|1),(1|2),(2|1),(2|2) \} \in \mathcal{P}(\mathbb{R}^2)$

$B= \{ \{(1|1),(1|2)\},\{(2|1),(2|2)\} \} \in \mathcal{P}(\mathcal{P}(\mathbb{R}^2)) $

If I sketch $A$ and $B$ on the plane, I get the same result.

Apologies if the question is unclear or relies on fundamental misunderstandings on my part.

I've read this thread: How to visualize $ \mathcal P \ ( \ \mathcal P \ ( \ \mathbb R ^2 \ ) ) $? but I still don't get it really.

LinusDieLinse
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    Your first mistake is trying to visualise these things. We have precise, and quite frankly, very simple definitions for these objects. Use them. That's it. With time you might develop intuition and an idea on how to visualise these sets, and that's great if you do, but it's not necessary or helpful. – Asaf Karagila Nov 28 '21 at 16:07
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    Examples of elements in $\mathcal{P}(\mathcal{P}(\mathbb{R}^2))$: The collection of Lebesgue measurable subsets of ${\mathbb R}^2,$ the collection of open subsets of ${\mathbb R}^2,$ the collection Borel subsets of ${\mathbb R}^2,$ the collection of curves in ${\mathbb R}^2$ that the Jordan curve theorem applies to, the collection of all circles in ${\mathbb R}^2,$ etc. – Dave L. Renfro Nov 28 '21 at 16:08
  • Before thinking about $\mathbb{R}^{2}$ it might be instructive to think about $\mathcal{P}(\mathcal{P}(\lbrace 0,1\rbrace))$. – Tobsn Nov 28 '21 at 16:15
  • @DaveL.Renfro - I like that you employed some measure theory in your explanation, I honestly never thought about this that way; thank you. – Taylor Rendon Nov 28 '21 at 17:23
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    Consider the cartesian plane as "model" of $\mathbb R^2$: a point is a pair of coordinates. Thus a line is a subset of $\mathbb R^2$, i.e. an element of $\mathcal P(\mathbb R^2)$. And then a set of lines is a subset of $\mathcal P(\mathbb R^2)$, i.e. an element of... – Mauro ALLEGRANZA Nov 28 '21 at 17:27
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    @Taylor Rendon: Thanks. Actually, I thought it was likely the measure theory and Borel set examples might be things the OP had not seen, but I left them in for others who might find them more "natural" than something like "the set of all circles in the plane". Also, when I first started typing the comment, I was going to mention $\sigma$-algebras of subsets of ${\mathbb R}^2,$ then before getting very far into doing this, I changed it to the less abstract and more specific examples of Lebesgue measurable subsets and Borel subsets. – Dave L. Renfro Nov 28 '21 at 20:20
  • Incidentally, Example #6 here exploits the cardinality distinction between $\mathcal{P}(\mathcal{P}(\mathbb{R}^n))$ (same cardinality for each positive integer $n)$ and $\mathcal{P}(\mathbb{R}^n)$ (same cardinality for each positive integer $n,$ which is also same as the cardinality of the collection of subsets of the boundary of the unit disk in the plane) for $n=1$ and $n=2.$ – Dave L. Renfro Nov 28 '21 at 20:27

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