1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

The set is

$$\mathcal{P}_f\left(\mathbb{R}^2\right)=\bigcup_{n=0}^{\infty}P_n\left(\mathbb{R}^2\right)$$

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 $n \in \mathbb{N}$).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.