In wikipedia I have found this about functions and the lexicographic order, it is basically the statement and I think it is a generalization of the lexicographic order.
Let's define the relationship $\preceq$ in $\mathbb{N}^\mathbb{N}$ as $f\preceq g$ if $f(n)= g(n)$ or if given $n$ the minimum natural number such that $f(n)\neq g(n)$ you have to $f(n)\leq g(n)$. Prove that $\preceq$ is a linear order in $\mathbb{N}^\mathbb{N}$
How can you prove that result? for the case of the lexicographic order it is easy to verify that it is a linear order, but for this case of the more abstract space I hope you can help me. Thanks in advance
Edit:
I have tried the following:
Reflexivity: Let $f\in \mathbb{N}^\mathbb{N}$, we have that $f(n) = f(n)$, then $f \preceq f$.
Antisymmetry: Let $f,g\in \mathbb{N}^\mathbb{N}$, if $f\preceq g$ and $g\preceq f$ it is trivial that $f = g$
Transitivity: Let $f,g,h\in \mathbb{N}^\mathbb{N}$ if $f\preceq g$ and $g\preceq h$, then $f=g$ and $g=h$, then $f=h$ or if given $n$ the minimum natural s.t $f(n)\neq g(n)$ and $g(n)\neq h(n)$ then $f\leq g$ and $g\leq h$ from this we can deduce that $f=h$ or if given $n$ the minimum natural number s.t. $f(n)\neq h(n)$ have $f\leq h$, then $f\preceq h$
It is right? And how can I see that the trichotomy property is fulfilled?
The given proof for transitivety has two mistakes.
Assume f <= g and g <= h.
Case f = g. Trivally f <= h.
Case g = h. Trivally f <= h.
Case f < g and g < h.
Exists smallest n with not f(n) = g(n)
and f(n) < g(n).
Exists smallest m with not g(m) = h(m)
and g(m) < h(m).
Case n <= m. Clearly f < h.
Case m < n. Show f < h.