# 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 $A$ ?

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.

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.

• You are confusing $A$ and the relation $S\subseteq A\times A$ on $A$. Under c) you talk about $1$ being an element of $S$. It is an element of $A$ but not of $S$. Feb 7, 2015 at 15:20

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 $$A=\{g\}=\{x\}$$
• a) looks okay. b) what makes you say: "there is at least 1 element which is not comparable to any other element in a set"? c) things go wrong here. E.g. all of a sudden you speak of a least element in $S$. Here $S$ is a (partial order) relation, but is not itself ordered. See also my comment on your question. Feb 7, 2015 at 16:06