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First part:

Prove that there's an infinite family $\mathscr{A}\subseteq P(\omega)$ such that:

  1. $X \in \mathscr{A} \Rightarrow |X|=\aleph_0$
  2. $(X,Y\in \mathscr{A} \wedge X\ne Y)\Rightarrow |X \cap Y|<\aleph_0$
  3. $\forall Z\subseteq \omega$, if $|Z|=\aleph_0$ then $\exists X\in \mathscr{A}$ such that $|X\cap Z|=\aleph_0$.

Second part:

Prove that $\mathscr{A}$ is not countable.

I've started the first part with Zorn's lemma, but when I get a promised maximum of the set $S=\{\mathscr{A}\subseteq P(\omega) : \mathscr{A}\ meets\ the\ 3\ conditions\}$ with the $\subseteq$ relation I can't know weather $\mathscr{A}$ is finite or infinite.

Maybe the solution is through [ultra]filters, compactness theorem or transfinite induction instead of Zorn's lemma. What do you think?

Andrés E. Caicedo
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J. Doe
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2 Answers2

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No, Zorn's lemma is plenty.

The trick, however, is to start with an uncountable family $\scr B$ satisfying (1) and (2), and define the partial order $S$ of all $\scr A$ such that $\scr B\subseteq A$ and the conditions hold.

However, we can also notice the following trend: either $\scr A$ is finite, e.g. if it is $\scr A=\{\omega\}$, or $\scr A$ is uncountable.

Suppose that $\scr A$ was countably infinite, say $\{A_n\mid n<\omega\}$, let $a_{n,m}$ be the increasing enumeration of $A_n$ for $m<\omega$. Then define $x_n$ to be the least $a_{n,m}$ such that $a_{n,m}\notin A_k$ for all $k<n$. Using the finite intersections, we can prove that such $m$ exists, so $x_n$ is well-defined.

Now $X\cap A_n$ must be finite, so condition (3) must fail.

Asaf Karagila
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  • For my understanding, I take (normalize the character style such that $A:= \mathscr{A},\ B:=\mathscr{B}$) $B:={X\subseteq \omega : |X|=\aleph_0 \wedge X \ is \ co-finite}$, then I define $S:={A:B\subseteq A}$. Then, what is my target $A$? Is it $A:=\bigcup S$? If so, the empty-set for example is inside $B\subseteq B\cup{\phi}$ which is inconsistent to (3). @AsafKaragila – J. Doe Jun 24 '18 at 20:02
  • The intersection of two cofinite sets is not finite. At least not when considering $\omega$. – Asaf Karagila Jun 24 '18 at 20:02
  • So how would you suggest to define$B$? – J. Doe Jun 24 '18 at 20:05
  • I don't really understand your goal here. Are you trying to tackle the proof in the second part of the answer? Try using an infinite partition of $\Bbb N$ (e.g. powers of primes and all the rest of the integers). – Asaf Karagila Jun 24 '18 at 20:07
  • No, I'm just in the first sentence of your answer; You suggested to take an uncountable family $B$ setassfying (1) and (2), so, I just try to figure out how to define $B$ explicitly. – J. Doe Jun 24 '18 at 20:10
  • https://math.stackexchange.com/q/162387/622 – Asaf Karagila Jun 24 '18 at 20:11
  • I got the idea that I can take $B$ that setassfies (1) and (2). Then you suggested to define $S:={A:B\subseteq A}$, then, what is the requiered $A\subseteq P(\omega)$ that settasfies (1), (2) and (3)? Do you mean $A:=\bigcup S$? I'm not sure what is the $A$ we want to find. – J. Doe Jun 24 '18 at 20:24
  • You wanted to apply Zorn's lemma, no? You need a partial order for that. – Asaf Karagila Jun 24 '18 at 20:25
  • Yes, but if I take $S:={A:B\subseteq A}$ then $B\cup{\phi}\in S$ doesn't setassfies the first condition. So with Zorn's lemma when I'll get the maximum (in the $\subseteq$ relation), I won't know for example weather the first condition is satsfied by my maximum. – J. Doe Jun 24 '18 at 20:31
  • If your partial order is made of families satisfying both (1) and (2), how can a maximal element in that partial order not satisfy both (1) and (2)? – Asaf Karagila Jun 24 '18 at 20:32
  • Oh my bad, in the end of the first part where you wrote "and the conditions hold" i thought you've find the $A$ and the conditions hold for that $A$, my misunderstood. – J. Doe Jun 24 '18 at 20:34
  • I defined $S:={A\supseteq B:\ A\ setassfies\ (1),\ (2)\ and\ (3)}$ but Zorn's lemma demands that $S\ne\phi$ and I can't know if there's any element in $S$. Moreover, I know that $B$ is infinte, so if I'll find 1 element in $S$, I'll get the $A$ that I want to prove its existance and I won't need Zorn's lemma. Where is my misunderstanding? – J. Doe Jun 24 '18 at 20:45
  • Why are you looking for all three conditions? When you're looking for a basis of a vector space, do you use the partial order of all maximally independent sets? – Asaf Karagila Jun 24 '18 at 20:51
  • I got it, what you're saying is that I get a maximal [changed from 'maximum'] $A$ form Zorn's leamma that setasfies (1) and (2). Assume in contradiction that $\exists Z\subseteq\omega\ s.t.\ |X\cap Z|<\aleph_0$ then, $A\cup{Z}$ setasfies (1) and (2) in contradition to the maximality of $A$. Thus, (3) is setasfied by $A$. – J. Doe Jun 24 '18 at 21:08
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    Yes. With the exception of "maximal" and not "maximum". – Asaf Karagila Jun 24 '18 at 21:08
  • We can start with $\scr B={n\cdot 2: n\in \omega}.$ Since $\scr A\supset \scr B$, this ensures that $\scr A \neq {\omega}$. – DanielWainfleet Jun 25 '18 at 06:03
  • @Daniel: Do you mean that $\scr B$ is a single set (in that case ${{n+n\mid n\in\omega}}$ would be the right notation). But then if you add the complement ${n+n+1\mid n\in\omega}$, then you have a partition and every infinite set meets one of them infinitely often. You need to start with infinitely many sets for that to work. Finitely many are just not enough. – Asaf Karagila Jun 25 '18 at 06:05
  • Sorry. You're right. Instead we can let $f:\Bbb Q\to \omega$ be a bijection and for each $r \in \Bbb R$ let $(q_{r,m}){m\in \Bbb N}$ be a strictly increasing sequence of rationals converging to $ r.$ And let $\scr B={ {f(q{r,m}) :m\in \Bbb N} : r\in \Bbb R}.$ – DanielWainfleet Jun 25 '18 at 06:13
  • @Daniel: That is what is in my answer in the question I linked in the comments. – Asaf Karagila Jun 25 '18 at 06:14
  • I must be too tired to see anything – DanielWainfleet Jun 25 '18 at 06:16
  • If I would like solving the question with tge help of compactness theorem/filters/transfinite induction, would it be possible? – J. Doe Jun 25 '18 at 06:43
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    @J.Doe: The last one is easy. Every use of Zorn's lemma can be replaced by a transfinite recursion, which essentially encapsulates the proof of Zorn's lemma. Enumerate $\mathcal P(\omega)$, and just add one set at a time until you have a maximal family as wanted, or you've exhausted all the sets. Compactness and filters, I'm not sure about that. These are weaker than full choice, and so their interactions with choice-related proofs is often harder. They also don't necessarily give you a "full picture", in the sense that ultrapowers/compactness arguments change your underlying "model". [...] – Asaf Karagila Jun 25 '18 at 06:51
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    [...] However you can appeal to Teichmüller–Tukey, which is a strengthening of the compactness theorem, and it will do the job. The proof is essentially the same as with Zorn's lemma, though. So I leave it to you to figure it out. – Asaf Karagila Jun 25 '18 at 06:52
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A topological argument: a family $\mathcal{A}$ that obeys 1. to 3. is called a maximal almost disjoint family in $\mathscr{P}(\omega)$. Almost disjoint means that two infinite sets have finite intersection. And the maximality follows from the last clause, as it implies that if $A$ is infinite and not in $\mathcal{A}$ it cannot be added to $\mathcal{A}$ without killing almost disjointness. Zorn applied the the poset of all such families (partially ordered by inclusion) yields, by standard arguments, the existence of an infinite such m.a.d.family (we merely extend any infinite pairwise disjoint family, say).

For any such a family we can construct a Mrówka $\Psi$-space denoted $\Psi(\mathcal{A})$ that has the following properties, proved from it being an infinite m.a.d. family (see here for construction and proofs):

  • $\Psi(\mathcal{A})$ has the form $\Psi(\mathcal{A})= \omega \cup \{x_A: A \in \mathcal{A} \}$
  • $\Psi(\mathcal{A})$ is locally compact, zero-dimensional, Hausdorff, separable, first-countable.
  • $\Psi(\mathcal{A})$ is pseudocompact but not countably compact (and thus not normal, as a normal psudocompact space is countably compact). $\mathcal{A}$ not being countably compact is shown by showing that $\{x_A: A \in \mathcal{A}\}$ is closed and discrete in $\Psi(\mathcal{A})$, and this set is infinite.

If a countable m.a.d. family $\mathcal{A}$ existed, then $\Psi(\mathcal{A})$ would be a countable, first countable (so second countable) regular space and hence metrisable and normal, which it is not. Ergo, no such family can exist. Quite indirect, but Mrówka's spaces are one of my favourite examples.

Henno Brandsma
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