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Here in the second comment I do not understand why $\omega^\omega$ corresponds to irrational numbers? : In my experience one typically identifies $ω^ω$ with the irrational elements of R; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".

QUESTION: What can we say about irrationality of $f=(2,2,2,2,...)$ goes to $(0,0,1,0,0,1,0,0,1,0,0,...)$ which is not irrational since it repeats the $0,0,1$ pattern forever,right? And yet, it doesn't end wiht eventually all $1$'s.

user122424
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    The usual correspondence maps elements of $\omega^\omega$ to infinite continued fractions, i.e., irrationals in $(0,1)$. It is not clear at all what the example you give is supposed to be. – Andrés E. Caicedo Jan 03 '19 at 16:40
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    The mapping that Noah Schweber gave in his post is not the one I was talking about. He does mention the one I was talking about in the sentence right before: it is the mapping that takes $(a_1, a_2, a_3, \ldots)$ to $\frac 1{a_1 + \frac1{a_2 + \frac1{a_3 + \cdots}}}$. – Mees de Vries Jan 03 '19 at 16:42
  • @MeesdeVries That explains my question, at least partially. However, how do I prove that all such $f's$ from $\omega$ to $\omega$ get map to irrationals via this continued fractions? What if there is a tail of $0$'s? Do they also map to irrational numbers? – user122424 Jan 03 '19 at 16:48
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    In particular, $(2, 2, 2, 2, 2, 2, \ldots)$ goes to an $x$ satisfying $x = \frac1{2 + x}$, so $x = \sqrt2 - 1$. – Mees de Vries Jan 03 '19 at 16:49
  • OK. But what about $(2,0,0,0,0,0,0....)$ ? – user122424 Jan 03 '19 at 16:50
  • Yes, they all map to irrational numbers. (My definition is slightly wrong if $0$ is a natural number: then you have to add 1 to each $a_i$ before forming the continued fraction.) In general, I recommend reading up on continued fractions/continued fraction expansions. – Mees de Vries Jan 03 '19 at 16:50
  • $\omega$ is an ordinal, am I the only one bothered by this mixing of notations between sets and ordinals when talking about mappings... Anyway, you can have a look at https://math.stackexchange.com/questions/2146599/encode-each-n-1-n-2-n-3-%e2%88%88nn-by-an-in%ef%ac%81nite-sequence-of-0s-and-1s-with-in%ef%ac%81ni/2146795#2146795 – zwim Jan 03 '19 at 16:59
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    @zwim The OP is using the notation correctly: ordinals are sets, and in particular $\omega$ is the set of all finite ordinals, and is generally used instead of "$\mathbb{N}$" in logic. – Noah Schweber Jan 03 '19 at 17:45

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As noted above, you're conflating two different homeomorphisms between $\omega^\omega$ and $\mathbb{R}\setminus\mathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.

Given an infinite sequence $A=(a_i)_{i\in\omega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1\over (1+a_1)+{1\over (1+a_2)+{1\over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):

$F_A$ is always defined and irrational and every irrational $\alpha$ is equal to $F_A$ for exactly one $A$.

See here for a proof. This implies immediately that the map $A\mapsto F_A$ is a bijection from $\omega^\omega$ to $\mathbb{R}\setminus \mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.

Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.

Noah Schweber
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    @zwim I don't think you read my answer carefully - I used $1+a_i$ instead of $a_i$ to deal with precisely this issue. E.g. taking $a_0=a_1=a_2=...=0$, we get $F_A=\varphi$. I believe my answer is correct as written. – Noah Schweber Jan 03 '19 at 20:29
  • Is there a continuous bijection between $\omega^\omega$ and entire $\mathbb R$? – user122424 Jan 19 '19 at 19:06
  • @user122424 Yes - indeed, if $\mathcal{X}$ is any "interesting" (= nonempty and perfect) Polish space, then there is a continuous bijection from $\omega^\omega$ to $\mathcal{X}$. I believe this was first proved by Sierpinski; it should be in Kechris' descriptive set theory book. Of course, order matters here - as I noted above, there is not even any nonconstant continuous map from $\mathbb{R}$ to $\omega^\omega$. – Noah Schweber Jan 19 '19 at 19:23