Find $A \subseteq \mathbb{R}$ which is ishomorphic to $\omega^3$

Question

I know by intuition that $\omega^{2}\cong\{k-\frac{1}{n}\mid k>0,n<\omega\}$

How can I find a set which is isomorphic to $\omega^3$? I would love to see an example If possible, I would like to see an example for $\omega^n$ or $\omega^\omega$ aswell.

Answer 1

You get $\omega$ embedded in $\Bbb Q$ (or $\Bbb R$) by just taking the natural numbers.

You get $\omega^2$ by taking some embedding of $\omega$, and in the gap between any two consecutive elements, insert a shrunk copy of $\omega$. For niceness (so that each multiple of $\omega$ in the result actually looks like a limit ordinal), make sure each copy "converges" to the element above it.

You get $\omega^3$ by taking some embedding of $\omega$, and in the gap between any two consecutive elements, insert a shrunk copy of $\omega^2$. For niceness, make sure each copy "converges" to the element above it. $$ \vdots $$ Continue this way to get any natural power of $\omega$.

To get $\omega^\omega$, start with some embedding of $\omega$ and in each gap between two consecutive elements, put a shrunk copy of higher and higher natural powers of $\omega$ (i.e. put $\omega$ between $0$ and $1$, put $\omega^2$ between $1$ and $2$, and so on). For niceness, make sure each copy "converges" to the element above it.

With some care, this can be generalized to embed any countable ordinal in $\Bbb Q$ (or $\Bbb R$).