0

Find an infinite collection of infinite sets $A_1,A_2,A_3\ldots$ such that $A_i\cap A_j=\emptyset$ for $i≠j$ and $$\bigcup_{i=1}^{\infty}A_i=\mathbb{N}.$$

My Attempt:

Let $p_i$ be the $i^{\textrm{th}}$ prime. For example, $p_1=2,p_3=5,$ and so on. Define a function on the natural numbers, $f$ such that $$f(n)=\textrm{the smallest prime that appears in the prime factorisation of }n.$$ For example, $f(2)=2,f(6)=2,f(15)=3,$ and so on. Now, let $A_i=\{n:f(n)=p_i\}.$ This gives us the partition we wanted. Each $A_i$ is infinite $(f(p_i)=f(p_i^2)=f(p_1^3)=\ldots).$ Of course, the intersection of two distinct sets is the empty set. It is not possible for a number to have two different minimum primes in it's factorisation. Moreover, each natural number is in some set, because each number has a prime factorisation and hence maps to some $p_i$ under $f.$

Does this look okay? What other ways can $\mathbb{N}$ be partitioned in? I was thinking we could also do something like $$A_i=\{n:n\textrm{ is a } i^{\textrm{th}} \textrm{ power and not a } j^{\textrm{th}} \textrm{ power for any } j>i\}.$$ In this case, $2\in A_1,64\in A_6,36\in A_2,$ and so on.

aqualubix
  • 1,794
  • 2
    $1$ isn't in your union, but you can fix that easily enough. You could also do something like the $2-$adic valuation (maximal power of $2$ dividing $n$. Or divide according to how many $0's$ appear in the decimal representation of $n$, and so on. – lulu Jul 08 '23 at 14:36
  • 1
    Looks pretty good to me! Another way of partitioning $\mathbb{N}$ (inspired by your alternative) would be to let $A_n$ be the set of perfect $n$-th powers, for $n\gt 1$. $A_1$ would be the set of natural numbers which are not a perfect $k$-th power for any $k\gt 1$. – H. sapiens rex Jul 08 '23 at 14:41
  • 1
    A simpler idea here: $A_n=2^{n-1}(2\mathbb N-1).$ Oh but @lulu already hinted that – Anne Bauval Jul 08 '23 at 14:59
  • @lulu, or we just put $1$ in $A_1$ and call it a day lol – aqualubix Jul 08 '23 at 15:01
  • 1
    @aqualubix if I understand the question correctly, each $A_n$ has to be an infinite set, doesn't it? $A_1$ would merely be finite were it to contain only 1... – H. sapiens rex Jul 08 '23 at 15:16
  • @H.sapiensrex, no I meant $A_1={n:f(n)=p_1}\cup {1}.$ – aqualubix Jul 08 '23 at 15:31
  • 1
    As I said, it is easy to fix the problem with $1$. – lulu Jul 08 '23 at 15:38

0 Answers0