# Subset of infinite partially-ordered powerset may not have maximal or minimal elements

I have recently been reading about partially-ordered sets and am practising some proofs regarding minimal and maximal elements in powersets. It all seems very intuitive when the powersets are finite, but I am not sure how to prove the nonexistence of a maximal or minimal element in an infinite set.

Suppose we have $$X \subseteq \mathcal{P}(\mathbb{N})$$

It seems fairly obvious to me that there should exist a non-empty subset X which does not have a maximal element because $\mathbb{N}$ is infinte. Likewise, there should also be an X which does not have a minimal element. But how does one prove these claims formally? Any guidance would be much appreciated.

Take$$X=\bigl\{\{1\},\{1,2\},\{1,2,3\},\ldots\bigr\}.$$Then $X$ has no maximal element. And if$$Y=\bigl\{\mathbb{N}\setminus\{1\},\mathbb{N}\setminus\{1,2\},\mathbb{N}\setminus\{1,2,3\},\ldots\bigr\},$$then $Y$ has no minimal element.

• Is $\mathbb{N}$ not the maximal element in $X$? – RexYuan Mar 1 '18 at 13:57
• Never mind my question - even if $\mathbb{N}$ is in $X$, $X - \{\mathbb{N}\}$ has no maximal element. – RexYuan Mar 1 '18 at 14:29

All the finite subsets of N have no maximal set.
All the subsets of N with finite complement have no minimal set.

Recall the following two facts:

1. If $A$ and $B$ have the same cardinality, then $\mathcal P(A)\cong\mathcal P(B)$ as ordered sets with $\subseteq$.

2. If $(A,\leq)$ is any partial order, then $(A,\leq)$ embeds into $(\mathcal P(A),\subseteq)$.

Now think about a countable linear order which has no minimal or maximal elements.