Toset - totally ordered set
For example, Take $(\mathbb{Z}, |)$ is not toset
Reason $(\{1,3,4,12\}, |)$ subset of above has hasse diagram not a chain
12
/ \
3 4
\ /
1
So with this can i conclude that $(\mathbb{Z}, |)$ is not toset? if subset not toset can i say set not toset? Are subsets of all tosets tosets? Converse? kindly elaborate.
The set-theoretic definition of a binary relation on a set $S$ is some (any) $R\subseteq S\times S.$ But we usually write $xRy$ instead of $(x,y)\in R.$
Let $R$ be a binary relation on $S$. Let $S'\subseteq S$ and let $R'=R\cap (S'\times S').$
(1). Suppose $p,q\in S'$ with $p\ne q$ such that $(p,q)\not\in R'$ and $(q,p)\not\in R'.$ That is, $p,q$ are $R'$-incomparable. Then neither $(p,q)$ nor $(q,p)$ belongs to $S$ because $S'\subseteq S,$ so $p,q$ are also $R$-incomparable.
If $p,q$ are unequal members of $S'$ and if exactly one member of $\{(p,q),(q,p)\}$ belongs to $R,$ then we also have $\{(p,q),(q,p)\}\subseteq S'\times S',$ so exactly one member of $\{(p,q),(q,p)\}$ belongs to $R\cap (S'\times S')=R'.$
If $p,q$ are unequal members of $S'$ and if exactly one member of $T=\{(p,q),(q,p)\}$ belongs to $R'$ then that member belongs to $R$ because $R'\subseteq R$. But the other member of $T$ cannot belong to $R,$ otherwise $T\subseteq R\cap (S'\times S')=R',$ which would imply both members of $T$ belong to $R'$.
(2). Similarly if $\forall x\in S\;(\,(x,x)\in R)$ (i.e. $R$ is reflexive) then $\forall x\in S'\;(\,(x,x)\in R')....$ OR if $x\in S$ but $(x,x)\not\in R$ then let $S'=\{x\},$ giving an $S'\subseteq S$ and an $x\in S'$ such that $(x,x)\not\in R'.$
In summary a binary relation $R$ on $S$ is total iff $R\cap (S'\times S')$ is total on every $S'\subseteq S$.
Tosets are also called linear orders.