# Are there $2^{\aleph_{0} }$ sets of natural numbers such that each two have finite intersection [duplicate]

Question: Are there $$2^{\aleph_{0} }$$ sets of natural numbers such that each two have finite intersection.

From what I've read about infinite families, I need to ignore those who have the properpty $$P$$.

Property $$P$$: the family is threadless, but whenever we take finitely many sets from the family, those sets have infinite intersection.

and probably find ones with property $$T$$.

Property $$T$$: the family is threadless, and it is an “almost-tower”: Whenever you pick two sets in the family, one of them is almost-contained in the other. - probably meaning that we can find such an intersection which is finite, because members are contained in each other.

Then I thought..

To find them, I should think of rational numbers instead of natural numbers, remembering that rational numbers can be paired up with natural ones, so solving this problem for families of rational numbers is the same as solving it for families of natural numbers. Now, I need to consider various ways of defining the real numbers.

And I'm stuck.. any help is appreciated.

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• What does "threadless" mean? Also, the families you are interested in are usually called "almost disjoint", this should let you google this. – Wojowu Apr 12 at 15:32
• Consider the set of sequences of rationals $A_r$, where $A_r$ converges to the real number $r$. – David Mitra Apr 12 at 15:35
• Zorn's lemma can be used to show there is a maximal such family. Not sure if that helps any. – Thomas Andrews Apr 12 at 15:39
• @DavidMitra Let’s say that two sets are almost disjoint if they only have finitely many numbers in common (or none at all). The set of primes is almost disjoint from the set of even numbers: they do have one number in common (who?), but that’s it. Also, can I just consider the set of numbers greater than a thousand which is almost disjoint from the set of numbers smaller than a trillion. There are plenty of numbers that appear in both sets, but only finitely many.. – Ilan Aizelman WS Apr 12 at 15:41
• I meant for $r$ to be irrational above. If $(a_n)$ and $(b_n)$ are sequences of rationals that converge to the irrational numbers $\alpha$ and $\beta$ respectively with $\alpha\ne\beta$, then $(a_n)$ and $(b_n)$ are almost disjoint. – David Mitra Apr 12 at 15:45

For $$\alpha$$ a real number, choose a single sequence of distinct rationals $$x_{\alpha}=(x_{\alpha,1},x_{\alpha,2},\dots)$$ that converge to $$\alpha.$$ Then the sets of elements in $$x_{\alpha}$$ and $$x_{\beta}$$ can only have finitely many rationals in common when $$\alpha\neq \beta.$$
So if $$X_{\alpha}=\{x_{\alpha,i}\mid i\in\mathbb N\}$$ then we have a family of size $$|\mathbb R|=2^{\aleph_0}.$$
If, in the case $$\alpha$$ is rational, we make sure $$\alpha\not\in X_{\alpha}$$, then we have that, for any $$\alpha,$$ $$X_{\alpha}$$ is a discrete bounded subspace of $$\mathbb Q$$ with $$\alpha$$ as the only limit point.
• Can $alpha$ be an irrational number in $[0,1]$? I'm asking because if it's irrational, then our sequence will be infinite - since we're picking from the range $[0,1]$ which makes the sequence not constant. with some more explanation i can say that because $\alpha != \beta$ then these sequence $x_{\alpha}$ and $x_{\beta}$ can have only finitely many terms in common since they have different limits. Meaning that their intersection is finite. – Ilan Aizelman WS Apr 12 at 15:59
• You don't need $\alpha$ irrational to assure the sequence is infinite. Every real $\alpha,$ rational or irrational, has an infinite sequence of distinct rational numbers that converges to $\alpha.$ Indeed if $\alpha$ is rational, we can define $x_{\alpha,n}=\alpha+\frac1n$ as a sequence of rationals. Not sure why you want to restrict yourself to $[0,1],$ either. I already said the sequence is non-constant (I said the elements were distinct.) – Thomas Andrews Apr 12 at 16:02