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.

  • 1
    $\begingroup$ 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 $S$. $\endgroup$ – drhab Feb 7 '15 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 $A=\{g\}=\{x\}$

  • $\begingroup$ Thanks, that means that yes, the same element can actually be both greatest and least, but could you tell me if my answers were correct ? $\endgroup$ – PainKiller Feb 7 '15 at 15:59
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    $\begingroup$ 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. $\endgroup$ – drhab Feb 7 '15 at 16:06
  • $\begingroup$ b)With the above conditions, (1 being both minimal and maximal), if it was comparable to any other element(other than itself), then it would either be minimal or maximal, it can't be both.c) to be honest, that's word to word what it says in the previous exam's papers, maybe there's a mistake. $\endgroup$ – PainKiller Feb 7 '15 at 16:22

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