Maximum and minimum of partially and totally ordered sets

Question

I have a problem I need help with: $M$ is an partially ordered set and $b\in M$ is a maximum if x $\le$ b and an element a $\in$ M is called minimum from M if a $\le$ x $\forall$ x $\in$ M. Prove that (i) If there is a maximum for the set M it is uniquely defined. Same goes for the minimum. What I have tried so far: Consider: a $\le$ x $\le$ b $\Rightarrow$ a $\le$ x $\Longleftrightarrow$ x $\ge$ a $\Rightarrow$ x $\subseteq a$ $\le$ b $\Longleftrightarrow$ x $\subseteq$ b $\ge$ a $$\\$$ So that (x,a) $\in$ R $\land$ (a,b) $\in$ R $\land$ (b,a) $\in$ R $\land$ (a,x) $\in$ R $\land$ (x,b) $\in$ R $\land$ (b,x) $\in$ R $\Rightarrow$ a = b because x $\subseteq$ a and x $\subseteq$ b. (Note that: R $\subseteq$ M$^2$). $$\\$$ (ii) If $M$ is also (a $\ge$ b or b $\le$ a) total. Then every finite, non-empty subset of $M$ has both a minimum and a maximum. I can see that I can probably use the mathematical induction but I do not see a proper way to solve this problem. I also have to prove whether (ii) applies to partially ordered sets. Any hints or solutions guiding to the right direction I much appreciate.

Answer 1

For point a), suppose both $a$ and $b$ are maximum. Then using the definition we know $a \geq x$ for every $x \in M$ so in particular $a \geq b$. In the same way, swapping $a$ and $b$, you prove $b \geq a$. This is possible only if $a=b$. The proof for the minimum is identical but with opposite inequalities.

For b) you can apply induction: let $A$ a finite subset of $M$. If $|A|=1$ then the maximum is the only element. Suppose now every subset of $M$ with cardinality $n$ has a maximum. Take $A$ of cardinality $n+1$ and take out an element $x$. The complementary in $A$ has cardinality $n$ so by hypothesis it has a maximum $y$. Now $x>y$ or $x<y$ as $M$ is totally ordered, so one of the two is the maximum (it is greater or equal than $x,y$ and so also of every other element of $A$ because $y$ was the maximum of the complement). For the minimum the same idea works.

For partially ordered sets it doesn't work: check that $a \leq b$ iff $a$ divides $b$ is a partial order on the set of natural numbers, but it has no maximal element (no natural number is a multiple of every other number).