1

I'm trying to build an increasing $\omega^{\omega}$-sequence on the real set but I'm not sure about it. Here is my approach.

Step $0$. consider the following $\omega$-sequence

$\langle\overbrace{\;0,\;1,\;2,\; 3,\ldots}^{\omega}\;\rangle$

Step $n+1$. Divide each unit of the n step increasing sequence in $\omega$ equal parts(this can be done because of the density of the real set).

In the case of $n=0$ we get an increasing $\omega^2$-sequence, specifically, this one: $\langle\overbrace{\underbrace{\;0,\;\frac{1}{2},\;\frac{2}{3},\;\frac{3}{4},\;\ldots\;,}_{\omega}\;\underbrace{1,\;\frac{3}{2},\;\frac{5}{3},\;\frac{7}{4},\;\ldots\;}_{\omega},\;2,\;\ldots\;,\;\omega}^{\omega\;natural\;numbers}\;\rangle$

And in the case of $n=1$ we get an increasing $\omega^3$-sequence, specifically, this one: $\langle\overbrace{\underbrace{\underbrace{\;0,\ldots,\;\frac{1}{2}}_{\omega},\underbrace{\;\ldots\;,\;\frac{2}{3}}_{\omega},\underbrace{\;\ldots\;,\;\frac{3}{4}}_{\omega},\;\ldots\;,1}_{\omega}\underbrace{\underbrace{\;\ldots\;,\;\frac{3}{2}}_{\omega},\underbrace{\;\ldots\;,\;\frac{5}{3}}_{\omega},\underbrace{\;\ldots\;,\;\frac{7}{4}}_{\omega},\;\ldots\;,2}_{\omega},\;\ldots\;,\;\omega\;}^{\omega\;natural\;numbers}\rangle$

Now, the question is: Would we get an increasing $\omega^{\omega}$-sequence in the real set at the $\omega$ step?

It can be done using recursion?

Thanks in advance for you help and time.

Marinovsky
  • 15
  • 2
    After $\omega$ steps your set will no longer be well-ordered. Let $x_n$ be the smallest positive member of the set at stage $n$; then $\langle x_n:n\in\omega\rangle$ is a strictly decreasing sequence after $\omega$ steps. – Brian M. Scott Jul 01 '20 at 19:23
  • 1
    For some approaches that do work, see https://math.stackexchange.com/a/123981 and https://math.stackexchange.com/a/336929 – Chris Culter Jul 01 '20 at 19:44
  • @Chris Culter: I just saw this (taking a quick break from day-job stuff) and thought maybe I'd link to something I remember posting that's relevant, assuming I could find it in a couple of minutes, but decided to click on your link first, and to my surprise your link goes straight to the answer of mine I was thinking of! – Dave L. Renfro Jul 01 '20 at 20:09
  • Hmm yeah, I haven't considered that sequence, thanks for the reply Mr. @Scott. – Marinovsky Jul 01 '20 at 21:29
  • Also thanks for the links @Chris Culter, especially, for the second one (I am still amazed of that idea). – Marinovsky Jul 01 '20 at 21:30

1 Answers1

3

Suppose for each $n\in \Bbb Z^+$ that $S_n\subset [n,n+1)$ and $S_n$ is order-iomorphic to $\omega^n.$ Then $\cup_{n\in \Bbb Z^+}S_n$ is order-isomorphic to $\omega^{\omega}.$

We can let $S_1=\{2-2^{-r}:r\in \Bbb Z^+\}.$

We can define $S_{n+1}$ recursively: For $m\in \Bbb Z^+$ let $T_{m,n}$ be order-isomorphic to $S_n$ with $T_{m,n}\subset [n+2- 2^{-m}, n+2-2^{-m-1}).$ And let $S_{n+1}=\cup_{m\in \Bbb Z^+}T_{m,n}.$

DanielWainfleet
  • 57,985