right now we're doing minimum/maximum/min/max element of sets and relations in class.
Now there's two questions I can't seem to find any idea on how to solve them.
Let M be an ordered Set (we write the relation as ≤)
(1)
Let M be a finite set and let x ∈ M be the only minimal element. Is x the minimum of M?
(2)
Let M be any set and x ∈ M be the only minimal element. Is x the minimum of M?
Now in class we've had the following:
If M is a partially order.
(i) Every minimum of M is minimal in M
(ii) If there exists a minimum in M, then it is the only minimal element of M. Particularly, that means that the minimum is unique.
Well problem (1) and (ii) looks like they're the same thing but (ii) assumes that there exists a minimum in M which isn't the case in (1) but my general idea is that (1) is a true statement and (2) is wrong because we could use an infinite set or just choose a set which has no minimum but one minimal element.
How exactly can you construct an example for (2) to dismiss the statement and how can you proof statement (1) by only using the definition of minimal element/minimum of a set?
Best Answer
Let $x$ in $M$ be the only minimal element and assume that $x$ is not the minimum. That means that there is some $y_0\in M$ such that $x\not\leq y_0$. However, also $y_0\not< x$ since $x$ is minimal. Thus $x$ and $y_0$ are incomparable. Now, $y_0$ is not minimal by the assumption that $x$ is the only minimal, so there exists some $y_1\in M$, $y_1<y_0$. Now, you can show that $y_1$ is not comparable to $x$ from how we picked $y_0$. Define $y_2$ analogously, and continue to construct a strictly decreasing sequence $(y_n)$. There goes finiteness out the window.
This should give you plenty idea for your second question. However, there is a common trick you can use is these problems. Take an ordered set that you understand well and add more elements to get what you need. In your case, take $\mathbb Z$. It has no minimal elements. Add a new element $*$ incomparable to any elements of $\mathbb Z$. It is obviously the only minimal element in $\mathbb Z\cup\{*\}$, but not the minimum.