Finding a poset with an antichain and failing pairwise join In an introductory exercise on posets and lattices, I am asked to give an example of a poset $(P,\leq)$ in which there are three elements $x,y,z$ s.t.


*

*$\{x,y,z\}$ is an antichain (EDIT: $A\subseteq P$ is an antichain if $a\not\leq b$ for distinct $a,b\in A$)

*$x\vee y, y\vee z$ and $z\vee x$ fail to exist

*$\bigvee\{x,y,z\}$ does exist


I have tried a lot of different things, including the standard $(\mathcal P(\{1,2,3\}),\subseteq)$, but everything I try comes after some time down to the same thing.
What am I missing?
 A: Here's some intuition: start with a discrete poset with three elements, $x$, $y$ and $z$. Add in a new element $\star = \bigvee \{ x,y,z \}$; this will be a single element which is greater than $x$, $y$ and $z$. If you leave it here then $x \vee y = y \vee z = x \vee z = \star$, so this doesn't satisfy your requirements. In fact, since $\bigvee \{ x, y, z \}$ exists, you'll need to add infinitely more elements to ensure that $x \vee y$ etc., don't exist. (Edit: this is a lie, see comments.)
To make sure that $x \vee y$ doesn't exist, add a copy of, say, $\mathbb{Q}$ (with its usual ordering), into your poset, such that each element of $\mathbb{Q}$ is greater than both $x$ and $y$, is incomparable with $z$, and is less than $\star$. Then do the same for the pairs $y,z$ and $x,z$. (You could use any poset with no least element instead of $\mathbb{Q}$.)
You can check that the resulting poset satisfies your requirements.

Putting this formally, define
$$P = \{ x, y, z \} \cup \mathbb{Q}_{xy} \cup \mathbb{Q}_{yz} \cup \mathbb{Q}_{xz} \cup \{ \star \}$$
where $\mathbb{Q}_{xy},\mathbb{Q}_{yz},\mathbb{Q}_{xz}$ are disjoint countable sets and the ordering $\le$ is the order generated by the following relations:


*

*$x \le t$ and $y \le t$ for each $t \in \mathbb{Q}_{xy}$;

*$y \le t$ and $z \le t$ for each $t \in \mathbb{Q}_{yz}$;

*$x \le t$ and $z \le t$ for each $t \in \mathbb{Q}_{xz}$;

*$\mathbb{Q}_{xy},\mathbb{Q}_{yz},\mathbb{Q}_{xz}$ are order-isomorphic to $\mathbb{Q}$;

*$t \le \star$ for each $t \in \mathbb{Q}_{xy} \cup \mathbb{Q}_{yz} \cup \mathbb{Q}_{xz}$.


There are some details to check but it's not too bad.
