ordering
Posted: Sat Jan 29, 2011 1:27 pm
Consider the following functions:
$g:\mathbb N \rightarrow \mathbb N; \ \ \ g(n) = $ highest power of $2$ that divides $n$
$f:\mathbb N \rightarrow P(\mathbb N); \ \ \ f(n) = $ set of all primes that divide $n$
Let $R_1$ and $R_2$ are two relations defined over $\mathbb N$ so that for any two members $a,b$ of $\mathbb N$, $(a,b) \in R_1$ if $g(a) < g(b)$ or, $a<b$ when $g(a) = g(b)$.
For $R_2$, $(a,b) \in R_2$ if $f(a) \subset f(b)$ or $a<b$ when $f(a) = f(b)$
Is there any bijection between $\mathbb N$ and $R_1$ or, $\mathbb N$ and $R_2$ or, $R_1$ and $R_2$
Apologis please if I've missed any mathematical rigor, I'm not that good in it
$g:\mathbb N \rightarrow \mathbb N; \ \ \ g(n) = $ highest power of $2$ that divides $n$
$f:\mathbb N \rightarrow P(\mathbb N); \ \ \ f(n) = $ set of all primes that divide $n$
Let $R_1$ and $R_2$ are two relations defined over $\mathbb N$ so that for any two members $a,b$ of $\mathbb N$, $(a,b) \in R_1$ if $g(a) < g(b)$ or, $a<b$ when $g(a) = g(b)$.
For $R_2$, $(a,b) \in R_2$ if $f(a) \subset f(b)$ or $a<b$ when $f(a) = f(b)$
Is there any bijection between $\mathbb N$ and $R_1$ or, $\mathbb N$ and $R_2$ or, $R_1$ and $R_2$
Apologis please if I've missed any mathematical rigor, I'm not that good in it