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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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Yes, your approach works.

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.