Linear continuum poset with succ and pred

Question

Is there a poset of continuum cardinality that satisfies conditions:

• it is linear
• there is a minimum element
• there is a maximum element
• each element (apart from maximum) has a successor
• each element (apart from minimum) has a predecessor ?

Answer 1

Yes! Initially, take your universe to be $\mathbb{Z}\times \mathbb{R}$ and order $(n,x)\leq (m,y)$ iff $x<y$ (normal order on the reals) or $x=y$ and $n\leq m$ (normal order on $\mathbb{Z}$). This gives you everything you want except the maximum and minimum.

So now throw in $\mathbb{N}\times\{\bot\} \cup \mathbb{Z}_{\leq 0}\times\{\top\}$. Put the $\mathbb{N}\times \{\bot\}$ part on the bottom, with the natural order, which gives the minimum while still having predecessors, and put the $\mathbb{Z}_{\leq 0}\times\{\top\}$ part on top with the natural order.

Alternatively, you can just start with $\mathbb{Z}\times[0,1]$ but remove $(n,0)$ for $n<0$ and $(n,1)$ for $n>0$, which achieves the same thing.

Edit: I feel I should add, the structure of the post actually reflects the way I constructed the example:

I saw continuum and thought $\mathbb{R}$ and needing predecessors and successors and thought $\mathbb{Z}$. So, I took the product and ordered them in a way that preserved the important properties of $\mathbb{Z}$. Then I needed to add a bottom, but then I had to add a successor to the bottom and a successor to that etc, which is something looking like a copy of $\mathbb{N}$, so I threw that in the bottom, and did something similar for the top element. Hopefully this help you construct examples in future.