We are given the sets $A=(1,2]\cup ((3,4)\cap \mathbb Q)$ and $B=(1,2)\cup ((3,4)\cap \mathbb Q)$ with the standard order $\leq$ of the reals.
Are they order-isomorphic? Meaning, is there a bijective function $f:A\to B$ such that $a_1 \leq a_2 \in A$ implies $f(a_1) \leq f(a_2) \in B$?
Answer: There isn't.
The reason for this (this is what the teacher said) is that the set $A^{*} = \{x\in A| |\{a\in A| a \geq x\}|\leq \aleph_0\}$ has a minimal value with the standard order. While $B^{*}=\{x\in B| |\{b\in B| b \geq x\}|\leq \aleph_0\}$ does not.
Firstly, I don't understand at all why this is true. And second, even if it is true, why does that imply that there isn't an order perserving isomorphism between $A$ and $B$? I don't see the relation between the 2 statements.
Answer
The difference between A and B is what happens with '2'. There is a hint, which is to prove that there is no order isomorphism, so we try to find something that is true in A and not true for B and that concerns orders.
Little digression as an example of what your teacher is trying to do : is there an homeomorphism between an infinite line and an infinite plane ? Answer : no. Because if you remove a point from a line you have 2 different connex sets, whereas this does not hold with a plane.
Here the teacher tries to find such a property (with sets that are countable/not countable) that should hold in both A and B if there was such an order isomorphism
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