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My question is pretty simple, but I want to make sure I'm understanding it correctly, because otherwise my misconception could go on a while.

What is the "order type" of [0,1)?

I know the set has a smallest element, does not have a largest element, and for any two elements in the set x < y we can find another element z so that x < z < y.

Are these the properties referred to by the phrase "order type of [0,1)"? Is there anything I am missing?

This was inspired by a homework problem from Munkres #12 from section 24 needing to show that [a,c) has the same order type as [0,1) iff both [a,b) and [b,c) have the same order type as [0,1).

Thanks!

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  • $\begingroup$ That does not capture it all. Two ordered sets have the same order type if they are order isomorphic. $\endgroup$ – André Nicolas Nov 14 '14 at 0:23
  • $\begingroup$ You miss just the fact that it has the supremum property. $\endgroup$ – Crostul Nov 14 '14 at 0:24
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Since the open interval $(0,1)$ is isomorphic to the real line, its order type is $\lambda$, the order type of the real line. The half-open interval $[0,1)$ therefore has order type $1+\lambda$; similarly, the order type of the half-open interval $(0,1]$ is $\lambda+1$, and the order type of the closed interval $[0,1]$ is $1+\lambda+1$.

I know the set has a smallest element, does not have a largest element, and for any two elements in the set x < y we can find another element z so that x < z < y.

Are these the properties referred to by the phrase "order type of [0,1)"?

No. The set $\mathbb Q\cap[0,1)$ of all rational numbers in $[0,1)$ has all the properties you listed, but is not isomorphic to $[0,1)$ (different cardinalities), so it does not have the same order type; its order type is $1+\eta$.

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  • $\begingroup$ Thank you! So then like @Andre Nicolas said I need to produce a bijection preserving order between the sets to show that they have the same order type? $\endgroup$ – sloth Nov 14 '14 at 0:37
  • $\begingroup$ @AlexH: That is the definition of having the same order type. In some settings there may be indirect ways to show that two linear orders have the same order type, but they are ways to show the existence of such a bijection without actually producing one. $\endgroup$ – Brian M. Scott Nov 14 '14 at 18:42

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