Give an example of a partially ordered set Give an example of a partially ordered set $(X,\le)$ satisfying $\forall x,y \in X,  \ \exists z\in X$ such that $x\le z$ and $ y\le z$ and $\forall x\in X ,\ \exists y\in X$ such that $x\not\le y$ and $y\not\le x$.
I understand I need to find a poset such that there is a maximum? and not all of the elements are comparable. 
I am thinking take $X=${$A_i|i\in \mathbb N$}$\cup \mathbb N$ such that $A_i=${$i$} and $\le$ is set inclusion. Then $A_i\subseteq\mathbb N \ \forall A_i$ and each of the singletons are not comparable.
I'd like some feedback for if this holds or how to improve it perhaps.
 A: A poset that satisfies the first condition need not have a maximum; that condition makes it a directed set. A directed set need not have a maximum element, and indeed the most interesting directed sets don’t have one. (I should mention that the order on a directed set can actually be just a preorder rather than a partial order.) Every linear order is a directed set; the second condition ensures that your ordering is far from linear.
Your example works, but your notation isn’t quite correct. You’ve made $\Bbb N$ a subset of $X$: the elements of your $X$ are the sets $A_i$ and the individual natural numbers $i$. What you wanted was $X=\{A_i:i\in\Bbb N\}\cup\{\Bbb N\}$, so that $\Bbb N\in X$.
A somewhat related example without a maximum element is obtained by letting $X$ be the family of non-empty finite subsets of $\Bbb N$, again with $\subseteq$ as the partial ordering. The union of two finite sets is finite, so for any $x,y\in X$ we have $x\cup y$ as an upper bound (in fact the least upper bound) for $x$ and $y$. The second condition is also satisfied: for any $x\in X$ there is certainly a $y\in X$ such that $x\cap y=\varnothing$, and clearly $x\nsubseteq y$ and $y\nsubseteq x$. (Alternatively, you could note that if $|x|=|y|$, then either $x=y$, or $x\nsubseteq y$ and $y\nsubseteq x$: distinct members of $X$ of the same cardinality are never comparable.) Note that I had to exclude $\varnothing$ from $X$, even though it’s a perfectly good finite subset of $\Bbb N$, because it is comparable with every subset of $\Bbb N$.
