How can you proof that there are $2^{\omega}$ perfect subsets of the real numbers?

  • $\begingroup$ If $\omega$ is the cardinality of the integers, then this is not the case. Or by $\omega$, do you mean the cardinality of the reals? If so, then how do you define $2^\omega$ anyway? $\endgroup$ – Omnomnomnom Aug 21 '13 at 19:21
  • $\begingroup$ @Omnomnomnom: $\omega$ is the cardinality of the integers, and the assertion is correct: there are $2^\omega$ perfect subsets of $\Bbb R$. $\endgroup$ – Brian M. Scott Aug 21 '13 at 19:28
  • $\begingroup$ 2$^\omega$ is defined as the cardinality of the reals (as is \omega the cardinality of the natural numbers) $\endgroup$ – ABC Aug 21 '13 at 19:29
  • $\begingroup$ @BrianM.Scott I misread the "perfect subsets" part of the question, and then I said something stupid. Thank you for the correction. $\endgroup$ – Omnomnomnom Aug 21 '13 at 19:47
  • $\begingroup$ @Omnomnomnom: I wondered if that’s what had happened. You’re welcome. $\endgroup$ – Brian M. Scott Aug 21 '13 at 19:49

Let $C$ be the middle-thirds Cantor set, and for each $x\in\Bbb R$ let $C+x=\{y+x:y\in C\}$ be the translate of $C$ by $x$. Clearly each $C+x$ is perfect, and $C+x\ne C+y$ if $x,y\in\Bbb R$ with $x\ne y$, so $\{C+x:x\in\Bbb R\}$ is a family of $2^\omega$ perfect subsets of $\Bbb R$. This shows that there are at least $2^\omega$ perfect subsets of $\Bbb R$.

By definition a perfect set is closed, so to finish the argument it suffices to show that $\Bbb R$ has only $2^\omega$ closed subsets. Since there is an obvious bijection between the closed sets in $\Bbb R$ and the open sets, we need only show that there are just $2^\omega$ open sets in $\Bbb R$. $\Bbb R$ has a countable base $\mathscr{B}$ (e.g., the family of open intervals with rational endpoints). For each open set $U\subseteq\Bbb R$ there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. There are only $\omega^\omega=2^\omega$ subsets of $\mathscr{B}$, so there are at most $2^\omega$ open sets in $\Bbb R$ and hence at most $2^\omega$ perfect sets.

Thus, there are exactly $2^\omega$ perfect subsets of $\Bbb R$.


There are at least $2^{\aleph_0}$ perfect sets, namely those of the form $[0,a]$, for $a$ a positive real. So it is enough to show that there are no more (by the Bernstein-Schroeder theorem).

But a perfect set is closed, and a closed set is the complement of an open set, so it is enough to check that there are at most $2^{\aleph_0}$ open sets.

Now, an open set is a countable union of open intervals with rational endpoints. There are countably many such intervals, so at most $\aleph_0^{\aleph_0}=2^{\aleph_0}$ open sets.

A very similar idea, coupled with something like transfinite induction, shows that in fact there are $2^{\aleph_0}$ Borel sets, see for example here.

Moreover: The Borel $\sigma$-algebra admits a natural stratification (or hierarchy) as $\bigcup_{0<\alpha<\omega_1}\mathbf\Sigma^0_\alpha$. Here, $\omega_1$ is the smallest uncountable ordinal, and:

  • $\mathbf\Sigma^0_1$ is the collection of open sets. Their complements (the closed sets) form the $\mathbf\Pi^0_1$ sets.
  • $\mathbf\Sigma^0_\alpha$ ($\alpha>1$) is the collection of countable unions of the form $\bigcup_{n=1}^\infty A_n$, where each $A_n$ belongs to some $\mathbf\Pi^0_{\beta_n}$ with $\beta_n<\alpha$. The complements of these sets form the $\mathbf\Pi^0_\alpha$ sets.

One can check that $\mathbf\Sigma^0_\alpha\subsetneq\mathbf\Sigma^0_\beta$ whenever $\alpha<\beta<\omega_1$. And a more refined argument than the one above shows that, for each $0<\alpha<\omega_1$, $|\mathbf\Sigma^0_\alpha\setminus\bigcup_{\beta<\alpha}\mathbf\Sigma^0_\beta|=2^{\aleph_0}$, and the same happens replacing each $\mathbf\Sigma$ with $\mathbf\Pi$. That is, it is not just that there are continuum many perfect sets: We also have that each level of the Borel hierarchy adds continuum many new sets.

(For a deep extension of this result in the choiceless context of determinacy, see here.)


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