@DHMO Hi

@AkivaWeinberger please begin your solution

So my construction used two facts:

that there are countably many finite sequences of $0$s and $1$s, and uncountably many infinite sequences of $0$s and $1$s

agreed

For any infinite sequence $s$, let $F_s$ be the set of all finite truncations

okay

$\{F_s\mid s\text{ is an infinite binary sequence}\}$ is thus an uncountable subset of $\mathcal P(\text{finite sequences})$ such that any two elements have finite intersection

We conclude by using a bijection between the set of finite sequences and $\Bbb N$ on each element of that set.

that is... amazing.

Another proof (that's essentially the same idea):

For each $s\in\mathcal P(\Bbb N)$, let $F_s=\{s\cap\{0,\dots,n\}:n\in\Bbb N\}$

and then we can use $\{F_s\mid s\in\mathcal P(\Bbb N)\}$

and a bijection between finite subsets of $\Bbb N$ and $\Bbb N$.

nice!

@AkivaWeinberger do you want to hear the one that I heard?

@DHMO Sure

firstly biject $\Bbb N$ with $\Bbb N^2$.

We will consider $\mathscr P(\Bbb N^2)$ instead.

OK

which are, you know, the lattice points of the first quadrant of the real plane

now consider the lines $y=mx$ with $m \in [0..\infty)$

there are $\mathfrak c$ many of them.

Ah!

and the rest is trivial.

I think I've heard this one, also, actually:

Do you take the ribbons of thickness $1$ around those lines?

And then intersect the ribbons with the lattice points

@DHMO

the one I heard isn't like this

actually not the lattice points

the unit squares.

intersect the lines with the unit squares

Oh.

to obtain a subset of $\mathscr P(\Bbb N^2)$.

Right. That's great!

Really clever.

indeed

but I don't like the fact that it uses $\mathscr P(\Bbb N^2)$

just like I don't like the fact that you use $2^\Bbb N$

I'm still looking a for a construction using $\mathscr P(\Bbb N)$ itself

@AkivaWeinberger the problem I like the most probably shouldn't be posted in this room

but I see no other suitable rooms

so let's go back to the main room

Hey there, everybody. I just have a little question.