Uncountably many sets of natural numbers with finite intersections [duplicate]

Possible Duplicate:
Countable set having uncountably many infinite subsets

Question:

Is it possible to find uncountably many infinite sets of natural numbers that any two of these sets have only finitely many common elements

You can even get $2^\omega=\mathfrak c$ of them. Such families of sets are called almost disjoint families.

One way to get an uncountable almost disjoint family of subsets of $\Bbb N$ is as follows. For each irrational number $x$ let $\langle q_x(k):k\in\Bbb N\rangle$ be a sequence of rational numbers converging monotonically to $x$. It’s easy to see that if $x$ and $y$ are distinct irrationals, the sequences $\langle q_x(k):k\in\Bbb N\rangle$ and $\langle q_y(k):k\in\Bbb N\rangle$ can have only finitely many terms in common. Now let $\varphi:\Bbb Q\to\Bbb N$ be any bijection, for each irrational $x$ let $D(x)=\{\varphi(q_x(k)):k\in\Bbb N\}$, and let $\mathscr{D}=\{D(x):x\in\Bbb R\setminus\Bbb Q\}$; $\mathscr{D}$ is an uncountable family of almost disjoint subsets of $\Bbb N$.

Added: Here’s another cute way. Imagine that you have an infinite plastic strip one unit wide. You pin some point on its midline to the origin, so that the strip can rotate about the origin. In each orientation it covers an infinite number of points of the integer lattice $\Bbb Z\times\Bbb Z$. For $\alpha\in[0,\pi)$ let $S_\alpha$ be the set of points in $\Bbb Z\times\Bbb Z$ covered by the strip when its midline coincides with the line $y=\alpha x$; it’s easy to see that $S\alpha\cap S_\beta$ is finite when $\alpha\ne\beta$. Now let $\varphi:\Bbb Z\times\Bbb Z\to\Bbb N$ be any injection, and for $\alpha\in[0,\pi)$ let $T_\alpha=\varphi[S_\alpha]$; the family $\{T_\alpha:\alpha\in[0,\pi)\}$ is then almost disjoint.