Given a number $n>1, n\in \mathbb N$, let $P$ be a totally ordered relation over set $A=\{1,2,..,n\}$. That is $(a,b)\in P \iff a\le b$.
In addition let $t\notin A$ and let relation $Q=\{(1,t),(t,t)\}$
Does $P\cup Q$ have a greatest and least element?
What troubles me is that we don't know what $t$ is, so maybe it's not even a number. In that case how can we determine what the least number is, by simply checking which element is in relation with all other elements? In our case it would be $1$ I guess.
But then if we don't know what is the exact relation between elements in $P\cup Q$ then we might have as well picked $1$ as the greatest element.
However if we suppose that $1$ indeed is the least element in $P\cup Q$ then in order to determine the greatest element I guess there're multiple cases:
- if $t\in \mathbb N$ then $t$ is the greatest element.
- if $t < 0$ then $t$ is the least element while $n$ becomes the greatest element.
- if $0\le t < 1$ - same as case 2.
- if $t$ is not a number then we don't have a greatest element because $t$ is not in relation with $A- \{1\}$
I'm not sure if I'm overcomplicating the problem with cases or whether I should only concentrate on case 4.
Your statement that "we don't know what is the exact relation between elements in P∪Q" is mistaken. We are told exactly what the relation is. You are correct that $1$ is the least element. $(1,a)\in P\cup Q$ because if $a\in A$ then $(1,a)\in P$ and if $a=t,$ then $(1,a)\in Q.$
There is no greatest element. The greatest element in $A$ is $n$, but $(t,n)\notin P\cup Q$
All that we are concerned with is $A\cup {t}.$ As William Elliot said in his answer, your remarks about numbers outside this set are off-base. In particular, it doesn't matter if $t$ is not a number. In fact, it's common to write $\infty$ to designate an element outside some set. We don't care what it is; we just know it's not in the set.