Are these two intervals order-isomorphic? Suppose $A (\text{unit}):= [0,1] \subset \mathbb{R}$ and $B:=[0,1] \cup [2,3]\, \cup...  \subset \mathbb{R}$. I am trying to show $A$ and $B$ might not be order isomorphic under the ordering of $<_\mathbb{R}$ on $\mathbb{R}$


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*I don't think A and B are order isomorphic under the condition $<_\mathbb{R}$ since they seem to not be equinumerous to one another. Such that there is not a bijection $A$ to $B$ that exists. Originally, I thought $f:\mathbb{R} \rightarrow \mathbb{R}: \tan(x)$ on the interval ($-\pi/2$, $\pi/2$) would be a bijective function, but scaling the interval accordingly to find this alternating union of distinct intervals in $B$ brought some issues. Any suggestions on how to move on from here?

 A: They are equinumerous. There is a bijection between them. So you can't use that to decide whether they are order isomorphic.
Hint: Are there any pairs of elements with nothing in between them?
A: Don't overthink this. Suppose there is an order-isomorphism $f:A\to B$. Then, there $0<x<y<1$ such that $f(x)=1$ and $f(y)=2.$ Then, $1<f\left(\frac{x+y}{2}\right)<2$, which is a impossible.
A: $A$ and $B$ are definitely equinumerous with one another. Neither are countable, and the notion that there are greater-than-countable subsets of $\mathbb R$ that are of any cardinality other than $|\mathbb R|$ is a little thing known as the Continuum Hypothesis, so you can pretty much assume that any non-countable subsets of $\mathbb R$ that you naturally come across are all equinumerous.
Generally to show that two structures $X$ and $Y$ cannot be isomorphic, you find some detail that $Y$ has which $X$ doesn't, and show that if one structure possesses such a detail, any structure isomorphic to it must as well. $A$, for instance, has no two elements which are "right next to each other", but $B$ does.
