# Indexed Families [duplicate]

I'm trying to find an indexed family {An:n∈N} that satisfies:

1. each An is an infinite subset of the naturals N
2. the intersection of any two arbitrary sets is empty
3. the union of all the subsets is is the naturals N

## marked as duplicate by Brian M. Scott elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 8 '16 at 20:28

• How about even numbers and odd numbers? – levap Jun 8 '16 at 20:05
• There must be many possible answers. What exactly was the particular difficulty that prevented you from finding one? – David K Jun 8 '16 at 20:13

Let $f\colon \Bbb N \times \Bbb N \to \Bbb N$ be any bijection, for example $$f(a,b) = \frac {(a+b)(a+b+1)} 2 + b.$$ Define $$A_n = \{f(n, k) \mid k \in \Bbb N\}, \quad \text{for n\in\Bbb N}.$$ Then the $A_n$ are pairwise disjoint and their union is $\Bbb N$. Every $A_n$ is infinite, because for each $n$, $k \mapsto f(n,k)\colon \Bbb N \to \Bbb N$ is an injection.
• Yes, every $A_n$ is infinite if $f$ is a bijection, because for each $n$, $k \mapsto f(n,k)$ is an injection. I'll add that to the answer. – BrianO Jun 8 '16 at 20:25
Define $A_n$ to be the set of all natural numbers on the $n$-th row. Needless to say that each $A_n$ is infinite.
There are as many ways of performing this task as there are orderings of $\mathbb{N}$.