Task: An element of order is called distant if it is greater than an infinite number of limit elements and less than an infinite number of limit elements.
a) Prove that with isomorphism of orders, distant elements pass into distant ones.
b) Give an example of an order in which there is a distant element that is not limit element.
Some clarifications: An element x of partial order is called limit element if it has no immediate predecessor.
My thoughts:
a) It seems that this point is proved by the definition of isomorphism. Let d be a distant element, then it is true that $d > (a_{1}, a_{2}.... a_{n}....)$ - these are some limiting elements. And $d < (b_{1}, b_{2}.... b_{n}....)$ these are other limiting elements. Then under isomorphism: $f(d) >f(a_{1})..., f(d) >f(a_{n})....$ and $f(d) <f(b_{1}) .... f(d) <f(b_{n})....$, as can be seen with isomorphism, the necessary inequalities that define the distant element are fulfilled. Is it possible to fully answer point a by presenting these arguments?
b) I can't come up with something normal: it is possible to take the ratio of the order $\leq$ all rational numbers and the element + $\infty$ then this element will be greater than infinite number of limit elements, but as for fulfilling the second condition: fewer infinite number of limit elements, difficulties arise...
We will work with $\mathbb{R}$ with the usual order. If we wanted to, we could work in $\mathbb{Q}$.
For each interval $[a,b)$, let the "convenient sequence on $[a,b)$" be given by $s_n=b-(b-a)2^{-n}$, $n=0,1,\ldots$. Then $a=s_0<s_1<\ldots$ and $\lim_n s_n=b$.
Pick $a,b\in \mathbb{R}$, a sequence $a=t_0<t_1<t_2<\ldots$ with $\lim_n t_n=b$. For each interval $I_n=[t_{n-1},t_n)$, let $(s^n_m)_{m=1}^\infty$ be the convenient sequence on $I_n$. Let $$T_1=\{s^n_m:m,n\in\mathbb{N}\}\cup \{b,b+1\}.$$
For each interval $J_n=[b+n,b+n+1)$, let $(r^n_m)_{m=1}^\infty$ be the convenient sequence on $J_n$. Let $T_2=\{r^n_m:m,n\in\mathbb{N}\}$. Let $T=T_1\cup T_2$.
Note that each $s^{n+1}_0$ is a limit element in $T$, since it is the limit of the elements $s^n_0<s^n_1<s^n_2<\ldots$. So $s^{n+1}_0$, $n=1,2,\ldots$ are infinitely many limit elements in $T$ which are less than $b+1$. The elements $b+2,b+3,b+4,\ldots$ are infinitely many limit elements in $T$ which are greater than $b+1$. But $b+1$ is not a limit element, because it has the immediate predecessor $b$. So $b+1$ is distant but not a limit element.
Or if you like ordinals, you can just take the poset $\omega^2\cdot 2$ and note that $\omega^2+1$ is distant but not a limit. The elements $\omega\cdot n$, $n=1,2,\ldots$ are infinitely many limits smaller than $\omega^2+1$, $\omega^2+\omega\cdot n$, $n=1,2,\ldots$ are infinitely many limits larger than $\omega^2+1$, and $\omega^2+1$ has $\omega^2$ as an immediate predecessor. But I believe this is actually order isomorphic to my example $T$.