I thought $[0,2]$ and $[0,1] \bigcup (2,3]$ are order isomorphic, since I can write the isomorphism $f:[0,2]\rightarrow [0,1] \bigcup (2,3]$ : $f(x)=x $ when $x\in [0,1]$
$f(x)=x+1$ when $x\in (1,2]$.
On the other hand, in $[0,2]$ every element has an immediate following element, and in $[0,1] \bigcup (2,3]$ the element 1 doesn't have an immediate following element.
I am a bit confused. Are they order isomorphic or not, and why?
Answer
No, in a real interval no number ever has an immediately following element. No matter which number you try to nominate as the immediately following element, there will be another one that is even closer.
$[0,2]$ and $[0,1]\cup(2,3]$ are indeed order isomorphic, for example by the correspondence you define.
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