Why does every non-empty subset of this poset have a supremum? I have a non-strict partial order $\preceq$ on $\{ 0, 1, 2 \}^{\mathbb{N}}$  given by 
$$f \preceq g \iff f^{-1}[\{ 2 \}] \subsetneq g^{-1}[\{ 2\}] \vee (f^{-1} [\{ 2 \}] = g^{-1}[\{ 2 \}] \wedge f^{-1}[\{ 1\}] \subseteq g^{-1}[\{ 1 \}])$$
My question is concerning a poset $<\{ 0, 1, 2 \}^{\mathbb{N}}, \preceq>$
Let $Z \subseteq \{0,1,2\}^{\mathbb{N}}$, $Z \neq \emptyset$
Why does $Z$ have a supremum? Intuitively I could just look at element of $Z$ with the biggest preimage of 2, if there are a few of these then look at the one with the biggest preimage of 1 and then create a sequence that has 2's on as many places as cardinality of preimage of 2 and 1's on as many places as cardinality of preimage of 1. That I think would work for finite number of 2's and 1's, but I think it doesn't work for sequences with infinite number of 2's or 1's,  for example a sequence with 2's on even indices, so I imagine this approach is flawed. How to approach this for these types of sequences?
 A: Let $Z = \{z_\alpha\}$ be s set of sequences and let each $z_\alpha$ be indexed so that $z_\alpha = \{z_{\alpha, k}|k\in \mathbb N\}$.
Define a sequence $w$ as:  if there exists any $z_{\alpha}\in Z$ so that $z_{\alpha,k} = 2$ then let $w_k = 2$.  Otherwise if exists any $z_{beta}\in Z$ so that $z_{beta,k} = 1$ then let $w_k=1$.  Otherwise let $w_k =0$.
Claim.  $w$ is a supremum of $Z$.
Pf:  For any $z\in Z$ if $k \in z^{-1}[2]$, that is if $z_k =2$ then we have defined $w_k = 2$ so $k \in w^{-1}[2]$.  So $z^{-1}[2]\subset w^{-1}[2]$. 
So either $z\preceq w$ or $z^{-1}[2]= w^{-1}[2]$.  If $z^{-1}[2]= w^{-1}[2]$ then if $j \in z^{-1}[1]$ then $z_j=1$.  Because $z^{-1}[2]= w^{-1}[2]$ and $j \not \in z^{-1}[2]$ we have $j\not \in w^{-1}[2]$ and $w_k\ne 2$. But $z_j=1$ so we defined $w_k = 1$.  So  $z^{-1}[1]\subset w^{-1}[1]$. And $z\preceq w$.
So either way $z\preceq w$.  So $w$ is an upper bound of $Z$.
Now let $u\in \{0,1,2\}^{\mathbb N}$ be so that $u\preceq w;u\ne w$.  If $u^{-1}[2]\subsetneq w^{-1}[2]$ then there is a $k$ so that $u_k \ne 2$ but $w_k =2$. BUt that means there is a $z\in Z$ so that $z_k =2$ and $u^{-1}[2]\not \subset z^{-1}[2]$ and $z \not \preceq u$.  So $u$ is not an upper bound.
If $u^{-1}[2]= w^{-1}[2]$ then $u^{-1}[1]\subseteq w^{-1}[1]$.  But $u\ne w$.  If $u^{-1}[1]-w^{-1}[1]$ and $u^{-1}[2]= w^{-1}[2]$ then $u^{-1}[0] = \mathbb N \setminus (u^{-1}[2]\cup u^{-1}[1]) =w^{-1}[0])$ and $u=w$.  So $u^{-1}[1]\subsetneq w^{-1}[1]$.
So there is a $k$ where $w_k =1$ but $u_k \ne 1$.  But $u\preceq w$ so $u_k \ne 2$.  So $u_k =0$.  But there is $z\in Z$ where $z_k =1$. And $z\preceq w$ so $z^{-1}[2]\subset w^{-1}[2]=u^{-1}[2]$  ..... Hmmmm I suppose it's possible that $z^{-1}[2]\subsetneq u^{-1}[2]$ and $u_k = 0$ and $z_k = w_k=1$ and still have $z \preceq u$.
Not sure how to finish this proof though it seems inuitively clear it'd be so... Som tedious induction should do it....
Don't know.  Leaving it open for now.  I'm fairly sure $w$ is the least upper bound of $Z$ though .....
