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A minimal element (any number thereof) of a partially ordered set $S$ is an element that is not greater than any other element in $S$.

The minimum (at most one) of a partially ordered set $S$ is an element that is less than or equal to any other element of $S$.

Let’s consider the power set $\mathcal P (\{x,y,z\})$ together with the binary relation $\subseteq$.

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The Hasse diagram shows what element(s) we’re looking for:

It’s easy to see that:

- $\emptyset$ is a minimal element
- $\emptyset$ is the minimum

Now if we remove $\emptyset$ and consider $\mathcal P (\{x,y,z\})\setminus \emptyset$ instead, we get the following:

- $\{x\}$, $\{y\}$ and $\{z\}$ are minimal elements
- there is no minimum

(1) We know that a minimum is unique and it is always the only minimal element.

(2) And from the example above, it seems that, if a sole minimal element exists, it is always the minimum.

But I read that (2) is false. Why?

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A counterexample to the statement is $\Bbb Z \cup \{c\}$, where $c$ is not comparable to any integer. $c$ is then a minimal element, but there is no minimum because a minimum must be comparable to everything else.

The poset suggested by the hasse diagram below has only one minimal element.

Generally, we could say:

If there is a unique minimal element, and every non-empty subset has a minimal element, then there is a minimum (and it equals the unique minimal element).

which is easily proven – suppose that in the poset $(S,\leq)$ our unique minimal element is $m$ and we consider $S’\subseteq S$ defined by $\{s\in S: m\not\leq s$} – so the elements incomparable with $m$. This cannot have a minimal element $m’$, since if it did, $m’$ would clearly be minimal in $(S,\leq)$ and distinct from $m$. However, this, in conjunction with the condition that every non-empty subset must have a minimal element implies that, since $S’$ has no minimal element, it empty, and thus $m$ is comparable with everything and hence a minimum.

However, this is a very strong condition – it is generally referred to as well-foundedness and means that no infinite descending chains may exist in the poset. The other answers provide explicit examples where there is an infinite descending chain, which precludes the possibility of a minimum element while having no minimal elements, which is in some way joined with a new minimal element.

A weaker but also sufficient condition would be

If there is a unique minimal element, and, for every subset $S’\subseteq S$, there is an element $m$ such that there is no $s’\in S$ with $s'<m$ and there exists an element $s’\in S$ such that $m\leq s’$, then $S$ has a minimum element.

which at least applies to posets like the typical order on $[0,\infty)$.

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