Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
"$[0,1)\times[0,1]$ is order-isomorphic to a subspace of the long line...." I'm not sure I follow how your outline works, but I'm very curious about it. I'll think about it some more after I get some sleep, but I'd love some more detail. –  Cameron Buie Sep 22 '12 at 7:57
    
I spoke too quickly on that, but since my intuition was of laying continuum-many closed intervals one after another, the fact that $[0,1]$ isn't well-ordered is going to kill the idea. –  Kevin Carlson Sep 22 '12 at 8:03
    
Maybe a way to make the order-isomorphy work would be to use intervals starting at the points of Cantor's discontinuum. –  celtschk Sep 22 '12 at 8:14
    
Is there an order-isomorphism between Cantor space and $[0,1)$? Seems like that's what's needed. –  Kevin Carlson Sep 22 '12 at 8:17
1  
With just a little more work you get that if $\langle A,\le_A\rangle$ and $\langle B,\le_B\rangle$ are linear continua, and $\preceq$ is the lexicographic order on $A\times B$, then $\langle A\times B,\preceq\rangle$ is a linear continuum iff $\langle B,\le_B\rangle$ has both a smallest and a largest element. And no, there is no order-isomorphism between the ternary Cantor set and $[0,1)$: the former has adjacent points and a right endpoint. –  Brian M. Scott Sep 22 '12 at 9:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.