My goal is to show that addition on well ordered sets are non-commutative by showing that,
$[0,1) +_o \mathbb{N} =_o \mathbb{N} \neq_o \mathbb{N} +_o [0,1)$
Some definitions (let A and B be posets)
$[n,m) =_{df} \{k\in \mathbb{N}: n \leq k \land k < m\}$
$A =_o B \Leftrightarrow_{df} \exists \ \phi : A \to B $ such that
$\phi$ is bijective and preserves order$A+_o B =_{df} \{0\}\times A \cup \{1\}\times B$ with $(i,x),(j,y) \in A+_o B$, $(i,x) \leq_{A+_o B} (j,y) \Leftrightarrow_{df} i < j \lor [i=j=0 \land x \leq_A y] \lor [i=j=1 \land x \leq_B y] $
I also take for granted (ie proved earlier) that sum of two wosets is a woset
I think I've found a order preserving bijection:
$\phi:[0,1) +_o \mathbb{N} \to\mathbb{N} $ by $\phi((0,0)) = 0,\ \phi((1,n)) = n+1$
$\phi(x) =\phi(y) \implies \phi(x) = 0 =\phi(y) = 0 \implies x = (0,0) = y$ or $\phi(x) =\phi(y) \implies \phi(x) = \phi((0,n)) = \phi((0,m)) \implies n + 1 = m+1 \implies x = y $ and in my mind it's clearly onto $\mathbb{N}$ since every natural number $n$ gets hit by either $(0,0)$ or $(1,n-1)$.
Order preserving.
Let $(i,n), (j,m) \in [0,1) +_o \mathbb{N} $ such that $(i,n) \leq_{[0,1) +_o \mathbb{N}} (j,m)$
Case 1.) $i< j$,
$(i,n) \leq_{[0,1) +_o \mathbb{N}} (j,m) \implies (i,n) = (0,0) \land (j,m) = (1,m) \implies \phi((i,n)) = \phi((0,0)) = 0 \leq_{\mathbb{N}} \phi((j,m)) = \phi((1,m)) = m+1 \geq_\mathbb{N} 1 $
Case 2.) $i = j = 0$
If $i = j = 0$ then $(i,n) \leq_{[0,1) +_o \mathbb{N}} (j,m) \implies (i,n) = (j,m) = (0,0) \implies \phi((i,n)) = \phi((j,m)) = 0 \implies \phi((i,n)) \leq_{\mathbb{N}}\phi((j,m)) $ Since $[0,1) = \{0\}$
Case 3.) $i = j = 1$
If $i = j = 0$ then $(i,n) =(1,n) \leq_{[0,1) +_o \mathbb{N}} (j,m)=(1,m)$ then we have that $n \leq_{\mathbb{N}} m \implies \phi((i,n)) = \phi((1,n)) = n+1 \leq_{\mathbb{N}} \phi((i,m)) = \phi((1,m)) = m+1 $
Here I draw the conclusion that $[0,1) +_o \mathbb{N} =_o \mathbb{N} $
But now to the part which I find very challenging. I have to show that $\mathbb{N} \neq_o \mathbb{N} +_o [0,1)$
I think I can find a bijection $\pi: \mathbb{N} \to \mathbb{N} +_o [0,1)$ by $\pi(0) = (1,0), \pi(n) = (0,n+1)$ but I can't get it to "break" with the order preserving. I've also tried to play around with just some general bijection but without any luck.
My idea is to find a bijection $\pi: \mathbb{N} \to \mathbb{N} +_o [0,1)$ (maybe a general one if that might work) assume or explicitly show that is has order. take two elements $a,b$ from $\mathbb{N} +_o [0,1)$ such that $a \leq_{\mathbb{N} +_o [0,1)} b$ and then map them to $\mathbb{N}$ thus we have $\pi(a),\pi(b) \in \mathbb{N} $ Since every element in $\mathbb{N}$ is mapped from some element in $[0,1) +_o \mathbb{N}$ we can rewrite $\pi(a) =\phi((i,n)),\pi(b) =\phi((j,m))$ and then deriving a contradiction so that $\pi$ can't preserve order. This at best a very very dirty sketch and I have several problems.
By intuition I refuse to believe that there is no bijective mapping from $\leq_{\mathbb{N} +_o [0,1)}$ to $\mathbb{N}$.
If I find a bijection and derive the contradiction, I have to show that this bijection is unique, right? I find this to be quite impossible to be honest.
Or am I going about this the wrong way, is it specifically the mapping of strict ordering that fails in some way I'm missing? Any help would be highly appreciated.
To show that $\mathbb{N} \neq_o \mathbb{N} +_o [0,1)$. We'll show by contradiction. Suppose that it exists an order preserving bijection $$\rho: \mathbb{N} \longrightarrow \mathbb{N} +_o [0,1).$$ Then, it must exist an element $m\in \mathbb{N} $ such that $\rho(m)=(1,0)$. But as $\mathbb{N}$ doesn't have a maximum element, we can affirm that $m+1$ exists. The image of $m+1$ would have the form $$\rho(m+1)=(0,x), x\in \mathbb{N}$$ and so $\rho(m+1)\leq_{\mathbb{N} +_o [0,1)}\rho(m)$, which is a contradiction because $\rho$ is order preserving.