Is $[0,1)\times[0,1]$ a linear continuum? I read that the topological space $X=[0,1)\times[0,1]$ with the dictionary order and order topology is not a linear continuum, as it does not satisfy the least upper bound property. (The definition of a linear continuum being a dense linear order with the least upper bound property.)
However, I can't find a nonempty bounded set with no supremum which leads to this violation. My thinking is if $A$ is any subset, and $\pi_1(A)$ is the projection onto the first coordinate, then $b=\sup(\pi_1(A))$ must exist, since if $\pi_1(A)$ is not bounded, then $A$ is not bounded above in $X$. Then the least upper bound of $A$ is $\sup(\pi_1(A))\times \sup(\pi_2(A\cap (b\times [0,1])))$. I feel the only difficulty occurs when $\sup(\pi_1(A))=1$, but then $A$ would not be bounded in the first place, so the situation doesn't apply. Is my thinking wrong, or is $X$ actually a linear continuum?
 A: Yep, this one is a linear continuum, thought your argumentation is not all flawless. If $b=\sup(\pi_1(A))=1$, then as you say
$A$ is unbounded; if $b<1$, then there are the cases $(1)$ that $A$ intersects $b=\sup(\pi_1(A))\times [0,1]$, and $(2)$, when it does not, as when $A=(0,1/2)\times[0,1]$.
In case $(1), \langle b,1\rangle$ is an obvious upper bound for $A$, so it's apparent that 
$\sup A=\langle b,\sup[\pi_2(A\cap(b\times [0,1]))]\rangle$ as you suggested. In case $(2)$, we must
still take $b$ the first coordinate of $\sup A$; but since no element of $b\times [0,1]$ intersects $A$ we get $\langle b,0\rangle$; so incidentally we could combine these two cases via the standard convention that $\sup \emptyset=\inf X$ for a linear order $X$.
Edit: bad idea removed.
You might have been thinking of $[0,1]\times[0,1)$, which isn't a linear continuum under the lexicographic ordering, because for instance $[0,1/2]\times [0,1)$ has $\langle 1,0\rangle$ as an upper bound but no least upper bound.
