Understanding the Order Types of n + ω and ω + n

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Discussion Overview

The discussion centers on the order types of the sets n + ω and ω + n, exploring why n + ω is order-isomorphic to ω while ω + n is not. Participants delve into the requisite isomorphisms and the implications of these order types, focusing on the theoretical aspects of ordinal numbers.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions the order type of n + ω and seeks clarification on the isomorphism that demonstrates it is equal to ω.
  • Another participant proposes an order-isomorphic representation of n + ω, suggesting a mapping T that relates elements of the set to ω.
  • A participant attempts to apply a similar approach to ω + n, questioning whether an isomorphism can be established and noting the challenges in doing so.
  • It is highlighted that in the case of ω + n, elements like (0,1) have an infinite number of predecessors, which contradicts the properties of ω, where each element has a finite number of predecessors.

Areas of Agreement / Disagreement

Participants generally agree on the isomorphism for n + ω and the inability to establish a similar isomorphism for ω + n. However, the discussion remains open regarding the implications and deeper understanding of these order types.

Contextual Notes

The discussion does not resolve the underlying assumptions about the definitions of order types and the nature of isomorphisms in this context.

Glinka
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Hi all,

Went over this today and I'm not grasping it: why is the order type of n + ω = ω, while ω + n ≠ ω? I'd really appreciate if someone could set up the requisite isomorphism in the former. Thanks!
 
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The set [itex]n+\omega[/itex] is essentially (order-isomorphic to) the following:

[tex](0,0)<(1,0)<(2,0)<(3,0)<...<(n-1,0)<(0,1)<(1,1)<(2,1)<(3,1)<...<(k,1)<...[/tex]

Do you see that??

The isomorphism between the above set and [itex]\omega[/itex] is given by the map T that does the following:

[tex]T(k,0)=k,~T(k,1)=n+k[/tex]
 
Ahh yes, this helps a lot, thanks. So in case of $$\omega + n $$ we could try $$(0,0) < (1,0) < ... < (k,0) < ... < (0,1) < (1,1) < ... < (n-1,1)$$ but we wouldn't be able to set up an isomorphism between this and ##\omega##?
 
Glinka said:
Ahh yes, this helps a lot, thanks. So in case of $$\omega + n $$ we could try $$(0,0) < (1,0) < ... < (k,0) < ... < (0,1) < (1,1) < ... < (n-1,1)$$ but we wouldn't be able to set up an isomorphism between this and ##\omega##?

Yeah exactly. Here we have the natural numbers and we paste n elements after it.
So take (0,1) for example. That has an infinite number of predecessors. So it can't be [itex]\omega[/itex] since any element in [itex]\omega[/itex] has a finite number of predecessors.
 
Excellent, thanks for your help!
 

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