Can the same element be greatest and least in a partially ordered set? I have an exam soon, so I'm just looking at the previous exams and solving every problem I see.
So the problem goes like this :
$R$ is a partially ordered set(relation) on $A = \{1, 2, 3, 4, 5\}$
$"1"$ is a minimal element in relation $R$ and $"1"$ is a maximal element in relation $R$.
a)Can you give an example of such relation on $A$ ?
My answer :
Let $R$ be a relation on $A$ that says :
$(x,y) \in R$ iff $x|y$ and $y|x$
Which means :
$R = \{(1,1), (2,2), (3,3), (4,4), (5,5)\}$
In this case $"1"$ is both minimal and maximal.
b)Give a general proof that if $R$ meets the above conditions, $R$ does not contain a greatest element and $R$ does not contain a least element.
My answer :
The greatest and the least element in a partially ordered set are comparable to every other element in a set(unlike minimal and maximal elements), if a relation meets the above conditions, there is at least 1 element which is not comparable to any other element in a set, which means it can't contain a greatest/least element.
c)This is the problem that is in the title, which goes like this :
Is there partially ordered set(relation) $S$ on $A = \{1,2,3,4,5\}$, such that $"1"$ is the Greatest element in S, and $"1"$ is the least element in S.
Give an example of such relation (if it exists), or prove that it doesn't.
Now, I can only think of a relation that goes :
$(x,y) \in S$ iff $x+y = 2$
In this case, the relation should contain only the element (1,1), which means 1 would be the minimal, maximal, greatest and the least element in $S$, is this correct ?Can the same element even be greatest and least at the same time ?
Any help is appreciated, would also be grateful if you told me whether the first 2 problems were answered correctly or not.

Let $(A,\leq)$ be a partial order and let it be that $x\in A$ serves as greatest element and as least element. Then $a\leq x\wedge x\leq a$ for each $a\in A$ and consequently $A=\{x\}$.
Let $(A,\leq)$ be a partial order and let it be that $x\in A$ serves as maximal element and as minimal element. If $g\in A$ serves as greatest element then $x\leq g$ and $x\neq g$ would contradict that $x$ is maximal. So we have $g=x$ so that $g$ is also minimal. As greatest element it is comparable with each $a\in A$ and the minimality of $g$ tells us that $g\leq a$ for each $a\in A$. That means that $g$ is a least element and going back to our former statement we find that $A=\{g\}=\{x\}$
