Order of minimun and maximum difference

Question

I want to prove firstly on $Fun(\mathbb{N},\mathbb{N})$ that "$f < g$ if and only if $f(k)<g(k), k=\min\{n|f(n) \neq g(n)\}$" is a total order, but not a well-order. Then, on the set $\{f:\mathbb{N_0} \to \mathbb{N_0}|f \ \text{is definitely} = 0\}$ "$f < g$ if and only if $f(k)<g(k), k=\max\{n|f(n) \neq g(n)\}"$ is a total and well-order. In the first case I can't find a counterexample to show that every non-empty subset has a least element in this ordering.
And, in the other case, how can I prove that the second set is well-ordered?

Answer 1

1. For all $k\in\mathbb{N}$, define $f_{k}(n)=\begin{cases} 0 & n\leq k\\ 1 & n>k \end{cases}$. Then $f_{k+1}<f_{k}$ since they coincide on $\{0,...,k\}$, but $f_{k+1}(k+1)=0<1=f_{k}(k+1)$. Thus, we have a descending infinite chain $f_{1}>f_{2}>...>f_{k}>f_{k+1}>...$.

2. For the second, Lets denote $D=\{f:\mathbb{N}_{0}\rightarrow\mathbb{N}_{0}|\exists n\in\mathbb{N}_{0},\forall k\ge n,f(k)=0\}$. Given $f\in D$, denote $n_{f}=min\{n\in\mathbb{N}_{0}|\forall k\ge n,f(k)=0\}$. For example, $n_{(0,0,...)}=0$, and $n_{(2,0,0,...)}=1$.

Lemma. Let $n\in\mathbb{N}_{0}$. Let $S'$ be a non-empty subset of $D$ such that for all $h\in S'$, $n_{h}\leq n$. Then $S'$ contains a least element $h_{0}$.

proof. By induction on $n$. For $n=0$, this is trivial, since $|S'|=1$ (why?). Assume $n>0$. Let $M=min\{h(n-1)|h\in S'\}$ (why $M$ exist?), and $S''=\{h\in S'|h(n-1)=M\}$ ($S''\neq\emptyset$). Note that if $g\in S'\backslash S''$, then for every $h\in S''$, $h<g$ (why?). So if $S''$ contains a least element, then this will be a least element of $S'$ also. For every $h\in S''$, we define $g_{h}(k)=\begin{cases} h(k) & k<n-1\\ 0 & otherwise \end{cases}$. Let $W=\{g_{h}|h\in S''\}$. $W$ is non-empty and $g_{h}\leq n-1$ for all $g_{h}\in W$. The induction hypothesis gives a least element $g_{h_{0}}$ of $W$. Finally, $h_{0}$ is the least element of $S''$ (why?).

$\blacksquare.$

Now let $S$ be a non-empty subset of $D$. Let $n=min\{n_{g}|g\in S\}$, and $S'=\{h\in S|n_{h}=n\}$ ($S'\neq\emptyset$). According to the lemma, $S'$ contains a least element $h_{0}$. We claim that $h_{0}$ is the least element of $S$. Assume $g\in S\backslash S'$. Then $n_{g}>n=n_{h}\geq0$, so that in order to compare $g$ and $h_{0}$ we have to compare $g(n_{g}-1)$ and $h_{0}(n_{g}-1)$ (why?). Of course, $h_{0}(n_{g}-1)=0<g(n_{g}-1)$, so $h_{0}<g$.

Answer 2

For the first question, consider the set $X:=\{f_n\in\omega^\omega:n\in\omega\}$ such that $f_n:=\chi_{[n,\infty)}$. This set hasn't a least element.

For the second, what means "$f$ definitely=$0$"?.