I'm exploring an unrelated question about power series with complex coefficients. While exploring this question, I wondered: What is the cardinality of the set of all such power series? Or with different language: What is the cardinality of at most countable subsets of $\mathbb{C}$ (or $\mathbb{R}$, if you prefer)?

I asked my advisor and he surprisingly wasn't sure, though he suspects that the set of subsets in question has a larger cardinality than $\mathbb{R}$.

Thanks a lot!

Edit: Certainly if we only consider finite subsets, then this set of subsets has cardinality equal to $\mathbb{R}$.

Edit2: Realized my wording was wrong. I'm actually looking for the cardinality of the set of sequences with entries in $\mathbb{C}$, not the cardinality of the set of at most countable subsets of $\mathbb{C}$. However, both questions are answered below, and both turn out to be $|\mathbb{R}|$.

  • $\begingroup$ Could the unrelated question be related to this 4 April 2013 comment of mine? $\endgroup$ Apr 21, 2014 at 20:57
  • $\begingroup$ This is exactly it. For a subset $S$ of the unit circle $T$ in $\mathbb{C}$ with finitely many limit points, I can construct a power series that diverges exactly at points of $S$ and nowhere else, using elementary stuff with the principal branch of log. I have also been wondering if there's an operation that takes a power series that diverges exactly at some given points of $T$ and outputs a power series that converges at exactly those points, and vice versa. No luck yet though. $\endgroup$
    – Dfrtbx
    Apr 21, 2014 at 22:03
  • $\begingroup$ Mickeysofine: $\Bbb C$ and $\Bbb R$ have the same cardinality. So it suffices to make the cardinality calculations for $\Bbb R$. $\endgroup$
    – Asaf Karagila
    Apr 22, 2014 at 16:26
  • $\begingroup$ I'm not so sure what you're getting at here, but I am aware. There are "fewer" power series than there are subsets of the unit circle. I'm just wondering what subsets correspond to power series that converge (or diverge, if you prefer) at exactly points of that subset. $\endgroup$
    – Dfrtbx
    Apr 23, 2014 at 13:35

2 Answers 2


(Note that $|\Bbb C|=|\Bbb R|$.)

If we assume the axiom of choice, then indeed the answer is that the set of countable subsets of $\Bbb R$ has the same cardinality as $\Bbb R$ itself.

The proof is simple. Using the axiom of choice, choose an enumeration of each such subset (i.e. a bijection between the countable set and $\Bbb N$). Next note the following cardinal arithmetic:

$$|\Bbb{R^N}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=|\Bbb R|.$$

Therefore the set of sequences of real numbers has the same cardinality as the real numbers. Now send each countable set to its enumeration, and we're done.

Without the axiom of choice, however, it is consistent that there are strictly more countable subsets. For example in Solovay's model where every set is Lebesgue measurable this is true.

  • $\begingroup$ Hmm... We need a site feature that tells you in real time who else is writing answers... $\endgroup$ Apr 21, 2014 at 16:45
  • $\begingroup$ @Andres: Yeah, although it's generally safe to assume that I'm writing one... ;-) $\endgroup$
    – Asaf Karagila
    Apr 21, 2014 at 16:46
  • $\begingroup$ To be honest, I don't quite understand why we need the axiom of choice here. If it is a countable subset, then of course we have at least one bijection between the countable set and N by definition (of countable). $\endgroup$
    – Michael
    Aug 20, 2022 at 9:46
  • 1
    $\begingroup$ @Michael Because you need to pick such enumeration for every countable set simultaneously. This is exactly where the axiom of choice is involved. Otherwise, your argument is that AC is provable: take a family of nonempty sets, then just pick one point from each set... $\endgroup$
    – Asaf Karagila
    Aug 20, 2022 at 10:31

The set of all countable subsets of $\mathbb R$ has size $\mathbb R$. This is because there is an injection from this set into the set $\mathbb R^{\mathbb N}$ of countable sequences. Now, $\mathbb R$ has the same size as $\{0,1\}^{\mathbb N}$ (for instance, see here), and we see that $|\mathbb R^{\mathbb N}|=|(\{0,1\}^{\mathbb N})^{\mathbb N}|=|\{0,1\}^{\mathbb N\times \mathbb N}|=|\{0,1\}^{\mathbb N}|$, since $\mathbb N$ and $\mathbb N^2$ have the same size.

Perhaps surprisingly, the answer depends on the axiom of choice. Tarski proved (without choice) that, for any set $X$, the collection of well-orderable subsets of $X$ has size strictly larger than $X$ (For a proof, see for instance here, or the references linked to here). It is consistent without choice that the only well-orderable subsets of $\mathbb R$ are countable (that is, that $\omega_1\not\le|\mathbb R|$). Whenever this situation holds, we have that the collection of countable subsets of $\mathbb R$ has size strictly larger than $\mathbb R$.


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