A family $\mathcal F$ of subsets of $\mathbb N$ is independent if for any two finite, disjoint subsets $\mathcal A,\mathcal B\subseteq\mathcal F$ the set $$\bigcap_{A\in\mathcal A}A\cap\bigcap_{B\in\mathcal B}(\mathbb N\setminus B)$$ is infinite.

It is well-known that there is an independent family on $\mathbb N$ of size $2^{\aleph_0}$. This for example implies that there are $2^{2^{\aleph_0}}$ ultrafilters on $\mathbb N$.

My favourite proof of the existence of a large independent family uses the Hewitt-Marczewski-Pondiczery Theorem that says that the space $2^{\mathbb R}$ (with the product topology) is separable: Pick a countable dense subset $D\subseteq 2^{\mathbb R}$ and consider, for each $r\in\mathbb R$, the set $A_r$ of all functions $f\in D$ (from $\mathbb R$ to $2$) with $f(r)=1$. The $A_r$ form an independent family of the required size on $D$.

There is a purely combinatorial proof as an exercise in Kunen's set theory book, but that proof is rather by computation than by visualization. There is a large number of nice proofs of the fact that there is a large almost disjoint family on $\mathbb N$.

So here is my question: Does anyone know a nice proof of the existence of a large independent family (large=of size continuum) on $\mathbb N$? (Other than the two proofs mentioned above or a combinatorialized version of the H.M.P.-argument.)

`$2^{\mathbb R}$`

with`$2^F$`

and let, for each $n\in\mathbb N$,`$f_n$`

be the element of`$2^F$`

defined by`$f_n(X)=1$`

iff $n\in X$. The set of these countably many`$f_n$`

's is dense, because when you untangle the definition of what it means for it to be dense, it boils down to the independence of $F$. $\endgroup$ – Andreas Blass Oct 7 '11 at 14:151more comment