Why is the order type of n + ω equal to ω but ω + n not equal to ω?

[Glinka]

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!

 

[micromass]

The set $n+\omega$ is essentially (order-isomorphic to) the following:

$$(0,0)<(1,0)<(2,0)<(3,0)<...<(n-1,0)<(0,1)<(1,1)<(2,1)<(3,1)<...<(k,1)<...$$

Do you see that??

The isomorphism between the above set and $\omega$ is given by the map T that does the following:

$$T(k,0)=k,~T(k,1)=n+k$$

 

[Glinka]

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 $\omega$?

 

[micromass]

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 $\omega$ since any element in $\omega$ has a finite number of predecessors.

 

[Glinka]

Excellent, thanks for your help!