I solved this question but there is something strange going on and I am unsure of myself. Would like someone to review it.
We are given a total order (or linear order) $<^{*}$on group $A$ such that for all $a \in A$:
$|\{x\in A | x<^{*}a\}|<\aleph_0$
We are asked to show that $A\leq \aleph_0$
My solution
if $<^{*}$ is a total order then there is an $a' \in A$ such that for all $x\in A$: $x<^{*}a'$.
so if we take $a=a': \{x\in A | x<^{*}a\}=A$ so we can infer that $A < \aleph_0$...
Here's what bothers me. We were asked to prove that $A \leq \aleph_0$ not $A < \aleph_0$. I know that what i proved shows both, but they aren't usually that lenient. if they wrote $\leq$ then that means there is a possiblity that $|A| = \aleph_0$, and according to my proof there isnt.
Have I done something wrong? I think that $A$ is a finite group. Else, $|\{x\in A | x<^{*}a\}|<\aleph_0$ won't be true for all $a$.
The issue here is this: Why does $A$ have a maximal element? It doesn't have to.
HINT: Show that if $|\{x\in A\mid x<a\}|=|\{x\in A\mid x<a'\}|$ then $a=a'$. Conclude that there is an obvious injection from $A$ into $\Bbb N$.