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Let $S$ be the set of all infinite subsets of $\mathbb N$ such that $S$ consists only of even numbers.

Is $S$ countable or uncountable?

I know that set $F$ of all finite subsets of $\mathbb N$ is countable but from that I am not able to deduce that $S$ is uncountable since it looks hard to find a bijection between $S$ and $P(\mathbb N)\setminus F$. Also I am not finding the way at the moment to find any bijection between $S$ and $[0,1]$ to show that $S$ is uncountable nor I can find any bijection between $S$ and $\mathbb N$ or $S$ and $\mathbb Q$ to show that it is countable. So I am thinking is there some clever way to show what is the cardinality of $S$ by avoiding bijectivity arguments?

So can you help me?

Dave L. Renfro
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4 Answers4

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Notice that by dividing by two, you get all infinite subsets of $\mathbb{N}$. Now to make a bijection from $]0,1]$ to this set, write real numbers in base two, and for each real, get the set of positions of $1$ in de binary expansion.

You have to write numbers of the form $\frac{n}{2^p}$ with infinitely many $1$ digits (they have two binary expansions, one finite, one infinite). Otherwise, the image of such a real by this application would not fall into the set of infinite sequences of integers (it would have only finitely many $1$).

Jean-Claude Arbaut
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The set of even numbers is countably infinite, and so equinumerous to $\Bbb N$ (explicit bijection given by $n\mapsto\frac12n$). With $F$ the set of finite subsets of $\Bbb N$, as you say, we can use the map $f(n)=\frac12n$ to get a bijection $g:S\to P(\Bbb N)\setminus F$, given by $$g(A)=\{f(n):n\in A\}.$$ I leave it to you to prove that it's bijective.

Cameron Buie
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  • So, if I understood your argument well, it is the case that if there is a bijection between $A$ and $B$ then there is a bijection between $R$ and $S$ where $R$ is the set of all infinite subsets of $A$ and $S$ is the set of all infinite subsets of $B$? –  Mar 25 '13 at 13:57
  • Absolutely right, and you can prove that as a general statement using the same idea that I outlined above. – Cameron Buie Mar 25 '13 at 13:59
  • Do you mean by the "general statement" that it works for every cardinality of $A$ and $B$ if we precisely enough define cardinality of $R$ and $S$, which is not needed in my case because every infinite subset in my problem must be of countable infinity? –  Mar 25 '13 at 14:05
  • What a flaw by me! It is fairly "obvious" what is the set of all infinite subsets of some set so in my previous comment maybe it is better to put away the word "precise". And you already answered that it is possible so it looks like I am asking the same question as in the first comment. It was a wave of confusion few minutes ago so it is clearer now for me to see it should work for every cardinality. Thank you! –  Mar 25 '13 at 14:14
  • Yes, even more generally, if $A$ and $B$ are in bijection, and if $C$ is some collection of cardinalities, then letting $R$ be the set of subsets of $A$ having one of the cardinalities in $C$, and letting $S$ be the set of subsets of $B$ having one of the cardinalities in $C$, we will have $R$ and $S$ in bijection. – Cameron Buie Mar 25 '13 at 15:24
  • When choosing cardinality for $R$ and $S$ from the some set of carinalities $C$ then we must choose same cardinality for both $R$ and $S$ because if not choosing the same then we couldn´t have a bijection between them because cardinality is defined in a way that sets have the same cardinality if there exists bijection between them, this is how I understood your statement, am I right? –  Mar 25 '13 at 15:36
  • Can you show that the collection of infinite subsets of the positive integers is uncountable? If so, then by doubling all the elements in each of these uncountably many subsets of the positive integers you will get uncountably many infinite subsets of the positive even integers. – Dave L. Renfro Mar 25 '13 at 16:38
  • @Thus: I should have made that more clear. I mean that $$R={X\subseteq A: |X|\in C}$$ and $$S={X\subseteq B:|X|\in C}.$$ There may be multiple cardinalities in $C$--for example, I could let $C$ be the class of finite cardinalities, finite even cardinalities, infinite cardinalities, uncountable cardinalities, etc. Does that make more sense now? – Cameron Buie Mar 25 '13 at 18:30
  • @DaveL.Renfro: That approach is basically what I did, only using $f^{-1}$ to induce the desired bijection. – Cameron Buie Mar 25 '13 at 18:32
  • @Cameron Buie: I was trying to strip away the formal explanations and technical terms to show the original poster what the essential idea is, in case the formalism in the answers was a problem. In looking over what he/she wrote a little more carefully, I'm now not sure what the problem is. The hard part is to show that there are at most countably many finite subsets. This forces uncountability of the infinite subsets, since $P({\mathbb N})$ is uncountable and equals the union of the collection of finite subsets with the collection of infinite subsets (so both collections can't be countable). – Dave L. Renfro Mar 25 '13 at 19:09
  • @Cameron Buie Do you know how to delete account here? Not that I am thinking of it because I like this place, but it seems strange that I do not see delete account button? –  Mar 27 '13 at 18:34
  • As soon as you voted or posted more than one question or answer, you were no longer able to delete your own account (though it was an option until that point). That's just StackExchange policy. If you ever do decide to delete your account for some reason, see here for instructions on how to request account deletion. – Cameron Buie Mar 27 '13 at 18:45
  • @Thus: Forgot to tag you in my response to your comment above. Apologies. Also, one final note on the generalization from above (with $A,B,R,S$). Suppose we have two sets $A,B$ of finite cardinality $n>0$, $f:A\to B$ a bijection, and $$R={X\subseteq A:|X|=1}\S={X\subseteq B:|X|=n-1}.$$ We can't induce a bijection $R\to S$ from $f$ in exactly the same way as I did in my answer, but $R$ and $S$ are still in bijection--define $g:R\to S$ by $$g(X)={f(x):x\in A\text{ and }x\notin X}.$$ (cont'd) – Cameron Buie Mar 27 '13 at 19:27
  • I suppose we could get even more general, yet, then. Suppose that $A$ and $B$ are in bijection, and that $C$ and $D$ are two collections of cardinalities such that $$W={X\subseteq A:|X|\in C}$$ and $$Z={X\subseteq A:|X|\in D}$$ are in bijection. Then $$R={X\subseteq A:|X|\in C}$$ and $$S={X\subseteq B:|X|\in D}$$ are in bijection, *even if* $C\cap D=\emptyset.$ In particular, let $f:A\to B$ and $h:W\to Z$ be bijections. Then $g:R\to S$ given by $$g(X)={f(x):x\in h(X)}$$ is a bijection. – Cameron Buie Mar 27 '13 at 19:36
  • This is becoming soooo general! :-D It is interesting, although intuition led me in the direction that it holds: $A$ and $B$ have the same cardinality if and only if the set of all subsets of $A$ have the same cardinality as the set of all subsets of $B$, that such a statement is unprovable, as I found out, and depends on the axiomatic approach to the set theory, did you know about that? –  Mar 27 '13 at 19:38
  • @Thus: I did know about that. That was one of the most surprising results I've run into, but choice principles (or lack thereof) make things pretty wonky sometimes when we're dealing with infinite sets. – Cameron Buie Mar 27 '13 at 19:40
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    So, if I ask you what approach would you choose for yourself, a set theory in which that statement is true, or set theory in which that statement is not true, what would you choose? You can choose it without an axiom of choice! ;.D –  Mar 27 '13 at 19:44
  • @Thus: Personally, I'd prefer for $A$ and $B$ to have the same cardinality if and only if their power sets do. This is a fairly weak choice principle, and I'd like it to be true for "aesthetic" reasons. – Cameron Buie Mar 27 '13 at 19:51
  • Could it be the case that Goldbach conjecture is undecidable, what I want to say is that could it be the case that we need to take that conjecture and put it in the theory of natural numbers as an axiom? Although it is for me hard to believe that it could be undecidable because surely it is the case that all even numbers greater than two are all expressible as sum of two primes or there are some that are not? What do you think? –  Mar 27 '13 at 20:01
  • @Thus: I'm afraid that's well outside of my area of expertise. I have no intuition about whether it is true, false, or undecidable. You might post that as a question, though. Somebody else might know if its decidability is...well...decided. ;-) – Cameron Buie Mar 27 '13 at 20:05
  • They do not want to tell me two trivial facts about Riemann zeta and surely that because of that they will choose not to discuss about the problem of decidability of Goldbach`s conjecture. Be aware that not all are friendly to others as you are to me. :-) –  Mar 27 '13 at 20:08
  • Thank you! What does it mean that the post is "community wiki"? –  Mar 27 '13 at 20:15
  • @Thus: I'd advise you to be patient, especially when you ask questions related to big undecided (though not necessarily undecidable) problems like the Riemann Hypothesis or the Goldbach Conjecture. The folks on this site (including myself) can be a grumpy and impatient lot, with varying degrees of frequency, and some folks would rather not even bother answering questions labeled "trivial". – Cameron Buie Mar 27 '13 at 20:15
  • @Thus: A Community Wiki answer is one that won't garner any reputation points for upvotes or acceptance. I chose to make that answer CW because I didn't really put any work into the answer (even though it's correct), and just posted it so that the number of unanswered questions would be reduced. – Cameron Buie Mar 27 '13 at 20:19
  • I understand what you´re saying to me, but I am almost sure that at least two persons who commented on my question about zeta know the answers and did not want to tell them, is it so hard to them to do what you did, write: "Yes and yes." and that is all I wanted to know. So if I ever go into the study of zeros of zeta then it is enough to prove the zeroes of Dirichlet eta are on the $\Re(s)=1/2$? –  Mar 27 '13 at 20:24
  • @Thus: Some folks would rather not post an answer that one could find on wikipedia (which shows that you're correct, by the way). I don't know why they often comment in such cases, but I suppose we're all peevish, sometimes. – Cameron Buie Mar 27 '13 at 20:38
  • So what would you recommend to me if I would like to set myself on the task of investigating the Riemann zeta or Dirichlet eta? Are there any materials available on the internet from which I can learn a lot? –  Mar 27 '13 at 20:56
  • @Thus: Once again, my knowledge of those two functions is fairly limited. I do know that much research has been done, though, and much of that is on the internet. You could ask that as a question with the (reference-request) tag, and someone more knowledgeable could point you in the right direction. – Cameron Buie Mar 27 '13 at 20:59
  • Good to know that such a tag exists! Maybe it is not a good move to ask such a question today because I already asked the question that has to do with those two functions on which you answered, so to ask two of such a questions one by another would maybe be a sign of alert. :-D So better if I ask it in few weeks or so. And thank you for chatting with me here! :-D –  Mar 27 '13 at 21:05
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Use the Cantor-Bernstein theorem, that is (in the simplified version)

$$|A| \leq |B| \text{ and } |B| \leq |A| \text{ implies } |A| = |B|.$$

We have that $S \subset P(\mathbb{N})$, so $|S| \leq |P(\mathbb{N})|$. Moreover, we have an injection $f : P(\mathbb{N}) \to S$ given by

$$ f(A) = \{4k \mid k \in A\} \cup \{4k+2 \mid k \in \mathbb{N}\}. $$

Hence, we have $|P(\mathbb{N})| \leq |S|$, so $|S| = |P(\mathbb{N})|$.

I hope this helps ;-)

dtldarek
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Hint: If $\{x_n\mid n\in\Bbb N\}$ is a set of even integers then $\{\frac{x_n}2\mid n\in\Bbb N\}$ is a sequence of integers. Show that this operation is in fact a bijection between $S$ and the set of all infinite subsets of integers. The latter is more familiar.

Asaf Karagila
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  • Thank you! I have a question that I can put in a comment and I believe that you could solve it, do you want of me to ask you? –  Mar 25 '13 at 14:28
  • If it is related, sure. – Asaf Karagila Mar 25 '13 at 14:29
  • I will answer it too, I just want of you to confirm is my line of reasoning correct. Let $S$ be the set of all subsets of $\mathbb N$ such that every set that belongs to $S$ consists only of numbers that are pairwise coprime (relatively prime). So I was thinking of this. The set of all prime numbers $P$ is in $S$ and it is infinite. Because the $P$ is infinite and every infinite subset of $P$ is also obviously consisted of coprime numbers then $S$ is uncountable because the $P$ is countable so it has uncountable number of countable subsets. Is this good reasoning? –  Mar 25 '13 at 14:43
  • If it is then I understood your arguments for the question I asked. –  Mar 25 '13 at 14:43
  • Yes, it is an excellent reasoning. – Asaf Karagila Mar 25 '13 at 14:48
  • Also, if you would be so nice to confirm to me if this is true, I am still an undergraduate so do not know enough math to confirm what my intuition leads me to. It seems to me that it should be true that $A$ and $B$ have the same cardinality if and only if the set of all subsets of $A$ have the same cardinality as the set of all subsets of $B$. Is this true? Surely this is trivial to you since I saw you´re Ph.D. student. –  Mar 25 '13 at 15:15
  • @Thus: The fact I am a Ph.D. student means nothing. But the answer is that while intuitively true, it is not true. It is not false either. It is unprovable. And by unprovable I mean that it is possible that there is an uncountable set whose power set is equipotent with the real numbers, like that of the natural numbers. But it is also possible to have a universe of set theory where that statement is true. Before you feel weirded out by this, remember that the answer to "how many fourth-roots does $1$ have?" also depends whether or not your universe includes the complex numbers or the reals... – Asaf Karagila Mar 25 '13 at 15:20
  • Surely I am weirded out! It sounds at the same time frightening and marvelous! Are you telling me that the truthness or falseness of the statement I gave depends on what axiomatic approach we choose when building the set theory? –  Mar 25 '13 at 15:27
  • @Thus: Yes. I do. I have written more than several answers on the topic, in particular referencing the problem of cardinality of power sets (which is one of the prime examples of an unprovable statement in set theory). But again, note that the existence of $\frac1x$ is not provable just from the axioms of addition. There is no $\frac12$ in $\Bbb N$. You need to assume more in order to have it, for example the axiom stating that multiplication has inverses. – Asaf Karagila Mar 25 '13 at 15:36
  • Thank you for this information you gave me here! Only I am left with a weird feeling of how someone can prove that something is not provable. Maybe if I investigate more there will be some place where I could find that there exists statement for which is unprovable that it is unprovable. Thank you again! –  Mar 25 '13 at 15:44
  • @Thus: If $T$ is a consistent theory (in first-order logic, whatever that means) then $T$ proves $\varphi$ if and only if in every model of $T$, $\varphi$ is true. In order to prove that $\varphi$ is unprovable one only needs to show that if $T$ has a model then there is one where $\varphi$ is true and another where $\varphi$ is false. Again from theory of field, $\exists x.x\cdot x=1+1$, saying that $\sqrt2$ exists. $\Bbb Q$ is a model of the theory in which this statement is false, $\Bbb R$ is another where it is true. Therefore we cannot prove $\sqrt 2$ exists from the axioms of the field. – Asaf Karagila Mar 25 '13 at 15:47
  • I have heard about Godel incompleteness theorem(s) and if you can explain me in few sentences what does those results of Godel show. I would appreciate if you could explain it by using as small number of technical terms as possible. –  Mar 25 '13 at 15:56
  • @Thus: These show that if we have a theory that a computer can recognize its axiom, amongst the sentences that we can write in the language, then either this theory is inconsistent; fails to prove some statement; or is too weak to "embed" the usual axioms of arithmetic (Peano axioms, and in fact less) into it. So in some sense, if a consistent theory can talk about natural numbers, and basic arithemtic, either it fails to prove something or we cannot recognize its axioms with a computer program. – Asaf Karagila Mar 25 '13 at 15:59
  • So you want to say that if we have $n_0$ axioms that we use to describe the theory of natural numbers there exists statement that we cannot prove using those $n_0$ axioms of natural numbers theory, so in order to prove it we must add additional axioms to natural number theory, in other words we need natural number theory with $m_0>n_0$ axioms to prove that statement, this look like a slight generalization of that which you explained to me, am I getting close to anything? –  Mar 25 '13 at 16:12
  • @Thus: Yes, that is correct. There are statements which are true in the standard model of the natural numbers (that is, the usual integers) that in order to prove we need to add additional axioms. Some nice ones include Goodstein's theorem; Paris-Harrington theorem; and the consistency of the theory itself. – Asaf Karagila Mar 25 '13 at 16:38
  • I would also like to know this: Is there a proof that in the standard model of natural numbers (or in some model of natural numbers that have additional axioms) there exists infinite number of statements that are not provable using those axioms? Godel showed that such statements exist but did someone been able to prove that there are infinite number of them? I mean countable infinity because it would be in some sense unappropriate to try to show that theory can have uncountable number of non-provable statements, that is almost beyond imagination. –  Mar 25 '13 at 16:53
  • @Thus: That much is trivial. If you have one unprovable statement $\varphi$, then $\psi\to\varphi$ is unprovable for every true $\psi$. But there can be more, for example the statement "Peano axioms are consistent" is unprovable from Peano axioms, but also the statement "Peano axioms + PA is consistent" is unprovable, and so is "Peano axioms + PA is consistent + (PA+PA is consistent) is consistent". And so on, and so forth. This gives us an infinite hierarchy of axioms, each stronger than the last. As for the last remark, finite/countable languages have only a countable set of sentences... – Asaf Karagila Mar 25 '13 at 16:57
  • Thank you very much for all of this! I am also aware that I am taking your time so I will stop with questions, your comments were very instructive and encourageable! :-D –  Mar 25 '13 at 17:01
  • Oh it's fine, I'm just waiting for the Seder to begin. That gives me another 10-15 minutes of loafing around... – Asaf Karagila Mar 25 '13 at 17:10
  • Celebrate it! I will delete the previous comment or this could turn into a discussion about Exodus! :-D –  Mar 25 '13 at 17:23