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Is there a metric space on $\omega^\omega$ such that $\alpha+n\to\alpha+\omega$ as $n\to\infty$?

Let $\omega^\omega$ be the set of all ordinals less than $\omega^\omega$ then I seek a function:

$d:\omega^\omega\times\omega^\omega\to\Bbb R$ such that $\omega^\omega,d$ is a metric space

and for all $\alpha\in\omega^\omega$, adding further integers converges to $\alpha+\omega$

I'm aware of the Order Topology but this looks to be far from a metric space.

it's a hire car baby
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  • You mean $\omega^\omega$ in the sense of ordinal exponentiation? – Brian M. Scott Jul 31 '20 at 19:51
  • @BrianM.Scott ‍♂️ I go through this every time I try to use ordinals. If I were to say cantor normal form, would they be a reasonable answer? – it's a hire car baby Jul 31 '20 at 19:53
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    I just wanted to make sure that you’re talking about the countable ordinal and not the cardinality of the countably infinite product of copies of $\omega$. – Brian M. Scott Jul 31 '20 at 19:55
  • @BrianM.Scott sounds more like the former than the latter. – it's a hire car baby Jul 31 '20 at 19:57
  • The order topology on any countable ordinal is metrisable (it's second-countable and normal and Hausdorff). – David Hartley Jul 31 '20 at 22:10
  • @DavidHartley I read about the order topology but couldn't find any metric on it. – it's a hire car baby Jul 31 '20 at 22:47
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    Urysohn's theorem tells you there is one, but to derive it directly from the proof of the theorem is not straightforward. Brian's answer gives you a much simpler construction. – David Hartley Aug 01 '20 at 10:40
  • @DavidHartley thanks. Does that say a space is "normal if its disjoint closed subsets can be separated by any continuous function", or "normal if they can be separated by some continuous function"? I fear this question may reveal the true degree of my ineptitude at topology! – it's a hire car baby Aug 01 '20 at 10:44
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    You're getting muddled with Urysohn's Lemma. Urysohn's metrisation theorem tells us that every second-countable, normal Hausdorff space is metrisable. ("some" not "any" in the lemma.) – David Hartley Aug 01 '20 at 11:42

2 Answers2

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Let $\alpha$ be any countably infinite ordinal. Fix a bijection $\varphi:\alpha\to\omega$ and define

$$f:\alpha\to\Bbb R:\eta\mapsto\sum_{\xi<\eta}2^{-\varphi(\xi)}\;;$$

then $f$ is an order-embedding of $\alpha$ in $\Bbb R$. If $\alpha$ has the order topology, $f$ is a homeomorphism of $\alpha$ onto $f[\alpha]$, and you can use it to define a metric on $\alpha$.

Brian M. Scott
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  • Here, $\alpha$ is my $\omega^\omega$, right? – it's a hire car baby Jul 31 '20 at 21:56
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    @samerivertwice: Yes, and this shows that it can be done with any countable ordinal. – Brian M. Scott Jul 31 '20 at 21:57
  • And an acceptable $\varphi$ would be to send the coefficients in cantor normal form to lengths of sequences of consecutive ones and zeros in an integer's binary representation. – it's a hire car baby Jul 31 '20 at 22:50
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    @samerivertwice: You couldn’t do that for arbitrary countable $\alpha$, because different countable ordinals can have the same coefficients but different powers of $\omega$. For $\omega^\omega$ you still have a problem, since $\omega^2+2$ and $\omega+2$ have the same non-zero coefficients, but you can get around it by using $c+1$ consecutive zeroes or ones to represent a coefficient $c$, so you can represent $\omega^3\cdot4+\omega\cdot2+1$, say, as $11111011100_{\text{two}}$. – Brian M. Scott Jul 31 '20 at 22:58
  • You say $f$ is an order-embedding of $\alpha$ in $\Bbb R$ but for some $\varphi$ what is to stop me defining another $\varphi'$ that's the same as $\varphi$ except for exchanging the chain $\omega\to\omega\cdot2$ with the chain $\omega\cdot2\to\omega\cdot3$? Then if one order-embeds, the other doesn't, no? (I ask more to correct my own misconception rather than to correct your claim!) – it's a hire car baby Aug 01 '20 at 12:49
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    @samerivertwice: No, you can use any bijection $\varphi:\alpha\to\omega$ and get an order-embedding; the real numbers $f(\eta)$ for $\eta<\alpha$ change if you change $\varphi$, but it’s still true that $f(\xi)<\f(\eta)$ iff $\xi<\eta$ and that $f$ preserves limits. – Brian M. Scott Aug 01 '20 at 17:02
  • Is this https://math.stackexchange.com/questions/3776055/is-there-a-metric-space-on-omega-omega-such-that-alphan-to-alpha-omega/3776065?noredirect=1#comment7774438_3776065 really a bijection because I'm having problems with $1,2,5,10,21,42\ldots\cong1_2,10_2,101_2,1010_2,10101_2,101010_2\ldots\mapsto \omega\cdot0,\omega\cdot0,\omega\cdot0,\omega\cdot0,\omega\cdot0,\omega\cdot0\ldots$? If this is a problem not easily solved I can ask a new question in its own right. – it's a hire car baby Aug 08 '20 at 10:13
  • It would seem a near solution may be $\varphi(\alpha)+\min{\beta\in\omega^\omega:\varphi(\alpha)+\beta=\beta}$. – it's a hire car baby Aug 08 '20 at 10:29
  • Actually scratch my last comment... currently favouring this as the bijection: $\omega\cdot\varphi(\alpha)+k$ where $k$ counts the nonempty coefficients $\alpha_i$ of $\varphi=\sum_{i\to k}\omega^i\cdot\alpha_i$ – it's a hire car baby Aug 08 '20 at 16:05
  • Your link seems to point to Angina Seng’s answer, so I’m not sure what map you’re talking about. – Brian M. Scott Aug 08 '20 at 17:01
  • Sorry, it points to your comment on this answer suggesting a bijection time stamped 22:58 on July 31st. – it's a hire car baby Aug 08 '20 at 17:17
  • @samerivertwice: It’s not a bijection, but it is an injection from $\omega^\omega$ to $\omega$, and that’s good enough for defining $f$. – Brian M. Scott Aug 08 '20 at 17:58
  • Am I mistaken in thinking $1,2,5,10,21,42\ldots\cong1_2,10_2,101_2,1010_2,10101_2,101010_2\ldots\mapsto0$ and therefore it's not an injection? Ah ok, it's an injection the other way, excluding $0$. Got it. – it's a hire car baby Aug 08 '20 at 18:07
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    @samerivertwice: The map goes in the other direction, from $\omega^\omega$ to $\omega$. The only one of those numbers that is in its range is $0$. – Brian M. Scott Aug 08 '20 at 18:10
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$\omega^\omega$ is a countable ordinal. One can embed every countable totally ordered set in an order-preserving way into $\Bbb Q$ and so into $\Bbb R$. Let's do this with $\omega^\omega$. But this need not be a homeomorphism of $\omega^\omega$ onto its image $X$ say. One can get round that by adding the upper bound of the image of every bounded increasing sequence in $\omega^\omega$. You'll then then a new subset $X'$ of $\Bbb R$ with the same order-type as $X$, and homeomorphic to $\omega^\omega$ in the order topology. Of course, $\Bbb R$ is a metric space, and so is $X'$.

You can do this for any countable ordinal.

it's a hire car baby
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Angina Seng
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  • Thank-you, much appreciated. – it's a hire car baby Aug 01 '20 at 12:41
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    @samerivertwice https://math.stackexchange.com/q/3722543/52912 has some details about the embedding procedure into $\Bbb Q$. – PatrickR Aug 01 '20 at 22:22
  • @PatrickR thanks Patrick I'll take a look. There is a little known embedding mentioned on this site which arose out of something I conjectured and Torsten Schoenberg proved which gives a straightforward bijection between $\omega^\omega$ and $\Bbb Q$. I just hadn't clicked that it had this convergence property. – it's a hire car baby Aug 01 '20 at 22:26