Subset of $\Bbb R$ isomorphic to $\omega+\omega$ and another isomorphic to $\omega\times\omega$

Question

I was trying to find two subsets:

• The first one is a subset of $\mathbb{R}$ that is isomorphic to $\omega+\omega$.

• The second one is a also a subset of $\mathbb{R}$ but isomorphic to $\omega\times\omega$.

Any suggestions?

Answer 1

• $\omega+\omega=\omega2\quad$ is equivalent to two infinite sets of increasing integers put end to end.

For instance take all even naturals followed by all odd integers. $$0,2,4,6,8,\cdots,1,3,5,7,\cdots$$

• $\omega\times\omega=\omega^2\quad$ is equivalent to $\omega$ infinite sets of increasing integers put end to end.

For instance take all the primes numbers, for each prime number consider all their powers, and so on. $$2^1,2^2,2^3,2^4,\cdots,3^1,3^2,3^3,3^4,\cdots,5^1,5^2,5^3,5^4,\cdots,\cdots$$

These are orderings on $\mathbb N$, but here the question is different, we want subsets of $\mathbb R$.

What is the difference ? In fact we can send the first list to $[0,1[$ then the second list to $[1,2[$, and so on... This way each infinity encountered on the natural sets can be sized back into an interval of $\mathbb R$ and the order induced by $<$ is now compatible with the desired ordinals.

For instance you can consider the increasing bijections $f_n(x)=n+\dfrac{x}{1+x}$ of $\mathbb R^+\leftrightarrow[n,n+1[$

For $\omega+\omega\quad A=f_0(2\mathbb N)\cup f_1(2\mathbb N+1)$ would do.

For $\omega\times\omega\quad B=\bigcup_\limits{p\in P} \{f_p(p^n)\mid n\in\mathbb N\}$ would also work.

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Although notice that we can find much simpler examples of subsets of $\mathbb R$.

In the examples above, we had to consider disjoint lists of natural numbers so they could form proper subsets of naturals, but since we send them to different intervals in $\mathbb R$, we could as well consider the same list for the problem at hand.

For instance what about $U_n=n+1-\{\frac 12,\frac 13,\frac 14,\cdots,\frac 1k,\cdots\}\subset[n,n+1[$

Note: I need to consider $n+1-\frac 1k$ and not $n+\frac 1k$ so that the order stays compatible with $<$ (else this would be with $>$).

$U_0$ is isomorphic to an infinite list of naturals with $\omega$ order.

Now considering unions of such sets, we have $U_0\cup U_1$ has $\omega2$ order and $\bigcup\limits_{n\in\mathbb N} U_n$ has $w^2$ order.

Remark: all these examples would work in $\mathbb Q$ too.