Which are linear continua? Definition of Linear continuum: A ordered set $L$ having more than one element is called linear continuum if te following holds:
(a) $L$ has least upper bound property.
(b) If $x<y$, there exists $z$ such that $x<z<y$.
The question:

Consider the following set in the dictionary order. Which are linear continua?
(1) $\mathbb{Z_+}\times [0,1)$
(2) $[0,1)\times \mathbb{Z_+}$
(3) $[0,1)\times[0,1]$
(4) $[0,1]\times[0,1)$

(1) is true since, for $x\neq y$, ($n\times x,m\times y$) always contains the points $n\times (\dfrac{x+y}{2})$ and $m\times (\dfrac{x+y}{2})$.
(2) is not true since there does not exists any point in $(x\times n,x\times \overline{n+1})$.
(3) is true, since $LUB$ do not need to be always inside $[0,1)\times[0,1]$.
(4) I could not find a solution. Ithink it is not true, but could not find an argument. Any help appreciated.
 A: In $[0,1]\times[0,1),$ what would be the least upper bound of the set $\{x\}\times[0,1),$ if $x<1\text{?}$ Every pair $(y,z)$ would be an upper bound of that set if $y>x.$ And for every such upper bound $((x+y)/2, z)$ would be a smaller upper bound. And there are no other upper bounds than pairs of that form.
A: Your statement in (3) that the $lub$ need not belong to $[0,1)\times [0,1]$ is invalid: Any $(L,<)$ is order-isomorphic to a subset of a linear order  $(L',<')$ such that every subset of $L'$ has a $lub$ in $L'$. So if we allow a $lub$ for a subset of $L$ to lie outside $L$ in the def'n of  "$lub$ property" then we would have to say that every linear order has the $lub$ property.
It is also not clear what is meant here by the $lub$ property:
(i). Every subset of $L$ has a $lub.$
OR (ii). Every non-empty subset of $L$ has a $lub$ in $L$.
OR (iii). Every subset of $L$ that has an upper bound in $L$ has a $lub$  in $L$.
OR (iv). Every non-empty subset of $L$ that has an upper bound in $L$  has a $lub$  in $L$.
