Suppose that $X$ is a poset with the following properties:

  1. $|X|=|\mathbb{R}|$,

  2. $X$ is totally ordered,

  3. $X$ has minimal and maximal elements (necessarily unique since $I$ is totally ordered).

Must there be a poset isomorphism $X\simeq [0,1]$? If not, what other properties should be added to the above list to provide such a characterization?

(Note that every real interval $[a,b]$, with $a\leq b$ real numbers, is isomorphic (as a poset) to the unit interval $[0,1]$ or the singleton $\{0\}$. So we can just find an isomorphism to some closed interval.)

It seems to me that one can define a poset representation to be an order-preserving map $\varphi:X\rightarrow \mathbb{R}$. For $\varphi$ to be injective, it is equivalent to be strict order-preserving (at the very least $X$ must have continuum cardinality), and so (3) should guarantee that the image of $\varphi$ is an interval.

So which other assumptions should be in place before one can one construct an injective order-preserving "representation" $\varphi:X\rightarrow \mathbb{R}$?


The conditions you have here are not sufficient to characterize the order type of an interval.

For an example of another order with all three of your properties, let $\kappa$ be the cardinal such that $\vert\mathbb{R}\vert = \kappa$, and let $X = \kappa+1$ (here I mean the next ordinal after $\kappa$). Then $\vert X\vert = \vert\kappa+1\vert = \kappa = \vert\mathbb{R}\vert$, so condition (1) is true. $\kappa+1$ is a well-order, so in particular is a total order, so condition (2) is true. Since $\kappa+1$ is a well-order it has a minimal element (namely $0$), and since it is a successor ordinal it also has a maximal element (namely $\kappa$), so condition (3) also holds. However, $\kappa+1 \not\cong [0, 1]$, because $\kappa+1$ is a well-order and $[0, 1]$ is not.

One property you definitely need $X$ to have is that the order on $X$ is dense, meaning that for every $a, b \in X$ with $a < b$ there exists $c \in X$ with $a < c < b$. Adding this condition to your (2) and (3) gives an axiomatization of the complete first-order theory of $[0, 1]$ in the language $\{<\}$. By standard results from model theory that theory does not have a unique model of cardinality $\vert\mathbb{R}\vert$ (if it did, then it would be uncountably categorical, hence stable, which it definitely isn't).

The upshot of the previous paragraph is that even if you add to your list everything that is true of $[0, 1]$ that is expressible in first-order logic in the language $\{<\}$, you still won't have characterized $[0, 1]$ up to isomorphism.

If you'd like a characterization of $[0, 1]$, there are some available. Perhaps the most well-known is that if $(X, <)$ is a linear order which is dense (in the sense I mentioned above), has a first and last endpoint, is complete (in the sense that every non-empty subset with an upper bound has a least upper bound), and has a countable dense subset (meaning there is some countable set $D \subseteq X$ such that for every $a, b \in X$ with $a < b$ there is $d \in D$ with $a < d < b$), and $\vert X \vert = \vert\mathbb{R}\vert$, then $X \cong [0, 1]$.

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  • $\begingroup$ Ah yes it makes a lot of sense that the completeness property should be there. Thanks! $\endgroup$ – Ehsaan Aug 7 '19 at 17:07

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