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Actually, I have tried the obvious fact that a subset of the line is connected iff it's an interval. And, the family of all possible intervals of the line is equinumerous with $\Bbb R$, as we can send any interval to an ordered pair having order as that of the end-points , including both the infinities.

But, for the plane, I am guessing about a suitable characteristic to match any connected set except the general definition. I thought the compliment, the cardinality of the family of sets which can be written as a separation, both non-empty, but then I think I need to find cardinality of all open subsets of the plane .

But, I need a hint to approach concretely .

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I am guessing about a suitable characteristic to match any connected set

Connected subsets of $\mathbb{R}^n$ for $n>1$ are waaaaaay too complicated to characterize. Unlike $\mathbb{R}$ case. For example for any countable subset $C\subseteq\mathbb{R}^n$ the subset $\mathbb{R}^n-C$ is even path connected. In particular surprisingly $\mathbb{R}^2-\mathbb{Q}^2$ is path connected! Somehow you can connect any two points while avoiding infinitely dense set of holes. Counterintuitive if you ask me.

But you don't need to know all connected subsets in order to solve the problem. Some is enough.

Consider $\mathbb{R}$ embedded into $\mathbb{R}^n$ onto first coordinate. When $n>1$ then there is $x\in\mathbb{R}^n$ which doesn't belong to $\mathbb{R}$. Now when $A\subseteq\mathbb{R}$ then construct a new set $A_x\subseteq\mathbb{R}^n$ by connecting every element of $A$ with $x$ via straight line. So $A_x$ is a cone. Note that $A_x=B_x$ if and only if $A=B$ and each $A_x$ is path connected (even when $A$ is not). Therefore there is at least $2^{\mathfrak c}$ (path)connected subsets of $\mathbb{R}^n$. And of course there cannot be more.

Note that this makes $\mathbb{R}^n$ for $n>1$ a lot more complicated than $\mathbb{R}$ which has only $\mathfrak c$ connected subsets.

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  • $\begingroup$ Ordered spaces are nicer; that's part of why $\Bbb R^n$ is so much harder to work with. In an ordered space its' easier to characterise connected subsets: no gaps and no jumps. $\endgroup$ – Henno Brandsma Jun 16 at 21:51
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$\Bbb R^2\setminus [0,1]\times \{0\}$ is connected, and it stays connected if you add in any collection of the points in $[0,1]\times \{0\}$. That gives you $2^{\mathfrak c}$ connected subsets, which is the same as all the subsets of $\Bbb R^2$. This works in any dimension greater than $1$

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  • $\begingroup$ And of course this generalizes to $\mathbb{R}^n$ $\endgroup$ – saulspatz Jun 16 at 19:33
  • $\begingroup$ @saulspatz any dimension, except $n=1$, to be precise (though I assume this is clear from the statement of the OP) $\endgroup$ – Mirko Jun 16 at 21:36
  • $\begingroup$ @Mirko Yes, of course. $\endgroup$ – saulspatz Jun 16 at 21:38
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For the plane, the answer is $2^\mathbb{R}$. For any subset $A\subset\mathbb R$, you can construct a different connected subset of the plane as $(A\times\mathbb R)\cup (\mathbb R\times\{0\})$.

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