How could I determine the cardinality of the set of all finite subsets of the plane? I believe I am correct in saying that this set is equivalent to the power set of $\Bbb R^{2}$ minus all infinite sets in that set. Is it correct to say that I can map any set of size $k$ to $\Bbb R^{2k}$, and then get a countable union of sets of cardinality $c$...which is $c$? That just sounds a little incomplete because how do we know this is a $countable$ union of said sets?

  • $\begingroup$ The cardinality of a finite subset is a natural number. There are countably many natural numbers, pretty much by definition. $\endgroup$ – Robert Israel Dec 15 '16 at 5:17
  • $\begingroup$ Actually you're mapping the sets of size $k$ to $\mathbb R^{2k}$, since each point has two coordinates. But the principle is the same. $\endgroup$ – Robert Israel Dec 15 '16 at 5:20
  • $\begingroup$ @RobertIsrael As for your first comment: I know that each set has finite cardinality, but what I meant was how do we know there is a countable union of those sets that are mapped to (the sets in $\Bbb R^{2k}$). What would an "uncountable" union even look like? $\endgroup$ – Wilson Brians Dec 15 '16 at 6:02
  • $\begingroup$ The set of all subsets would be an uncountable union. (This is commonly called the Power set) $\endgroup$ – b00n heT Dec 15 '16 at 6:04

The set is


where $P_n\left(\mathbb{R}^2\right)=\{S\subset\mathbb{R}^2\,;\,|S|=n\}\subset{\left(\mathbb{R}^2\right)}^n$.

Now, clearly, the cardinality of $P_1\left(\mathbb{R}^2\right)$ is $\left|\mathbb{R}^2\right|=|\mathbb{R}|$. For $n> 1$, the cardinality of $P_n\left(\mathbb{R}^2\right)$ is also $|\mathbb{R}|$, because there is a trivial injection onto $\mathbb{R}^n$. At this point it suffices to use this theorem on cardinality of infinite unions to deduce that $\left|\mathcal{P}_f\left(\mathbb{R}^2\right)\right|=|\mathbb{R}|$.

Notice that basically the same argument yields that if $A$ is any infinite set and $\mathcal{P}_f(A)$ is the set of finite subsets of $A$, then $|\mathcal{P}_f(A)|=|A|$. Here we used the handy fact that for any infinite cardinal $\kappa$ it holds that $\kappa\times\kappa=\kappa$ (which implies that $\left|A^n\right|=|A|$ for all $n \in \mathbb{N}$).

  • $\begingroup$ right, I should have said "the power set of $\Bbb R^{2}$ minus infinite sets in that set" in my original question. $\endgroup$ – Wilson Brians Dec 15 '16 at 16:21

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