# Non-trivial order on a set with infinitely many minimals/maximals

I am looking for an example of each:

1) A non-trivial order on a set A such that there are infinitely many minimal elements

and

2) A non-trivial order on a set A such that there are infinitely many maximal elements.

I was thinking that A could bet the set of ℝ. For a,b∈A, a≤b iff a+b. I don't know if this is trivial or not because it is true for both; there are infinitely many minimals or maximals. But I need two different examples.

"Order" meaning a relation that is reflexive, transitive, and anti-symmetric.

All examples are welcome, mathematical or not. Thank you!

Here is one example. Let $A$ be the set of integers greater than $1$. Define $x$ to be less than or equal to $y$ if $x$ divides $y$.

For example, $6$ is "less than or equal" to $42$, but $11$ is not.

The minimal elements are the primes. There are infinitely many of them.

To get infinitely many maximals, turn this order upside down.

Here is another example. Let $A$ be the collection of non-empty subsets of the natural numbers. If $x$ and $y$ are such subsets, write $x$ is less than or equal to $y$ if $x\subseteq y$. The $1$-element sets are the minimal elements.

Perhaps you can use these two to "roll your own."

Let $X$ be an infinite set, consider $\mathcal P(X)$ ordered by inclusion. It has a minimum and maximum, but what happens when those are removed?

Now the minimal elements are those which are singletons, and how many of them do we have? The maximal elements are those subsets missing just one element, and again -- how many of these do we have?

Take any two disjoint, infinite sets $X$ and $Y$. Let $A=X\cup Y$. Define the order to be $$\{(p,q): p=q\text{ or }(p\in X\text{ and }q\in Y)\}.$$ The set of minimal elements is $X$ and the set of maximal elements is $Y$.