A linear continuum is a totally ordered set with more than one element, which is both dense and satisfies the least upper bound property. Is it true that every linear continuum has the same cardinality as the reals?
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No, there are larger linear continua. Let $\kappa$ be any infinite cardinal; then $X=\kappa\times[0,1)$ with the lexicographic order is a linear continuum. It’s easily seen to be densely ordered. If $A\subseteq X$ is non-empty and bounded above by some $\langle\alpha,x\rangle$, let
$$\beta=\min\left\{\gamma<\kappa:\big(\{\gamma\}\times[0,1)\big)\cap A=\varnothing\right\}\;.$$
If $\beta$ is a limit ordinal, then $\sup A=\langle\beta,0\rangle$. If $\beta=\gamma+1$ for some $\gamma<\kappa$, let $$S=\{y\in[0,1):\langle\gamma,y\rangle\in A\}\;.$$ If $\sup S=1$, then again $\sup A=\langle\beta,0\rangle$. Otherwise, $\sup A=\langle\gamma,\sup S\rangle$.
answered Dec 11, 2013 at 7:33
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$\begingroup$ @BrianMScott: So what extra axioms must one add to those for a linear continuum to insure that its models are all isomorphic to $\mathbb R$? $\endgroup$ Mar 12, 2017 at 6:28
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$\begingroup$ @Sushil: (I must not have been around when you posted your question.) This is very late, but for the benefit of future readers, the answer is no. Every countable dense linear order is order-isomorphic to $\Bbb Q$, $\Bbb Q\cap[0,1)$, $\Bbb Q\cap(0,1]$, or $\Bbb Q\cap[0,1]$, none of which has the least upper bound property. $\endgroup$ Jun 14, 2022 at 19:00