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!
Answer
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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