Let $A=[0,1)\times [0,1]$ have dictionary topology. Does it have the least upper bound property?
My answer: Yes, any element of the form $B=\{1 \times m: m\in [0,1]\}$ can be the least upper bound. Am I correct?
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The set $A$ itself does not have an upper bound in the dictionary order. Are you asking whether it has the least upper bound property, meaning that every subset of $A$ that has an upper bound has a least upper bound? – Brian M. Scott Oct 21 '13 at 02:28
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@BrianM.Scott: Yes, Brian that is what I meant. Sorry for the confusion – James Bond Oct 21 '13 at 02:31
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You’re still asking whether $A$ itself has a least upper bound, and I’m pretty sure that this is not the question that you want to ask. I’m going to edit the question to make it what I think you want to ask; after I’ve done that, please tell me whether it’s correct or not. – Brian M. Scott Oct 21 '13 at 02:38
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@BrianM.Scott: I see what you meant before. I wasn't aware of the difference. Thank you for stating it. And yes, the edited question is precisely what I want to ask – James Bond Oct 21 '13 at 02:41
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See also: [Is $0,1)\times[0,1]$ a linear continuum? – Martin Sleziak Oct 20 '17 at 14:43
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This linear order does have the least upper bound property, but what you’ve written doesn’t prove it. You need to start with an arbitrary non-empty $S\subseteq A$ that has an upper bound $u=\langle x_0,y_0\rangle\in A$ and show that $S$ has a least upper bound in $A$.
HINT: Let $S_0=\{x\in[0,1):S\cap(\{x\}\times[0,1])\ne\varnothing\}$. Show that $S_0$ has an upper bound in $[0,1)$ and conclude that it has a least upper bound $s_0$. Then let $S_1=\{y\in[0,1]:\langle s_0,y\rangle\in S\}$, show that $S_1$ has a least upper bound $s_1\in[0,1]$, and finally prove that $\langle s_0,s_1\rangle$ is the least upper bound of $S$ in $A$.
Brian M. Scott
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