# Minimals in partially ordered sets

In a subset $Y$ of a partially ordered set $(X, \leq)$ a minimal element is an element $a \in Y$ such that $(\forall y \in Y)\ y \leq a \Leftrightarrow y = a$

However in this diagram (http://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Hasse_diagram_of_powerset_of_3_no_greatest_or_least.svg/435px-Hasse_diagram_of_powerset_of_3_no_greatest_or_least.svg.png) the elements $\{x\}, \{y\}$ and $\{z\}$ are all minimals. Shouldn't this mean that $\{x\} = \{y\} = \{z\}$, according to the above definition?

No, because $(x,y),(x,z),(y,z)$ are not in the relation. In other words, you dont have $x\leq y$ so you don't have $x=y$.

Its easier to talk about partially ordered sets (posets) rather than subsets thereof. So lets phrase the definition for posets.

Definition. Suppose that $P$ is a poset and that $k \in P$ is an element. Then $k$ is minimal iff the following holds. $$\forall p \in P (p \leq k \rightarrow p = k)$$

(The $\rightarrow$ could easily be replaced by a $\leftrightarrow$ and the definition would be the same.)

Now suppose $P$ is a poset and that $x$ and $y$ are minimal in $P$. Then we may deduce that

$$\forall p \in P (p \leq x \rightarrow p = x), \qquad \forall p \in P (p \leq y \rightarrow p = y)$$

Hence:

$$(y \leq x \rightarrow y = x), \qquad (x \leq y \rightarrow x = y)$$

So if either one of $\{x,y\}$ is less than or equal to the other, then they're equal. But this may not be the case. For example, consider the poset with two elements $\{x,y\}$ such that the statements $x \leq x$ and $y \leq y$ are true, but the statements $x \leq y$ and $y \leq x$ are false.