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$A\cup B=C$

Suppose some function $f$ totally orders equivalence classes of $A$ and equivalence classes of $B$, both of which are formed by the same equivalence relation $\sim$.

The domain and range of $f$ is $C$

The range of $f$ in $A$ is $A$

The range of $f$ in $B$ is $B$

Does this imply that $f$ totally orders equivalence classes of $C$ formed by the same equivalence relation?

2 Answers 2

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For a counter-example let $C=\{1,2,3,4\} , A=\{1,2\} $ and $B=\{3,4\}.$ Let the set of equivalence classes be $E= \{\{1,3\} , \{2,4\}\}.$

By "the equivalence classes of $A$" I assume you mean $\{x\cap A: x\in E\}$ \ $\{\phi\}.$ So the equivalence classes of $A$ are $\{1\}$ and $\{2\}.$ And the equivalence classes of $B$ are $\{3\}$ and $\{4\}.$

Suppose $\{1\}<\{2\}<\{4\}<\{3\}.$ This doesn't "translate" to a linear order on $E.$

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    But $f$ must totally order equivalence classes of e.g. $A$ while the range of $f$ in $A$ is $A$ so it cannot state $\{3,1\}$ are equivalent. – 2017-01-24
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    Sorry I clarified that in the notes earlier; I should have added to the question - I will now. My bad. – 2017-01-24
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    Do you mean that an equivalence class of C is always a subset of A or of B? – 2017-01-24
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    yes I think that's the case because there is no $f(b)=a$ nor $f(a)=b$ – 2017-01-24
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Not necessarily. We could have $f$ map an equivalence class from $A$ and a different equivalence class from $B$ to the same element in your order and that would break the totality.

If $f$ is one-to-one or $f|_A$ and $f|_B$ have disjoint ranges, that would give a total order.

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    The case i have in mind, $f$ is many to 1 but the domain and range of $f$ in $A$ is $A$ and the domain and range of $f$ in $B$ is $B$. Is that what $F\lvert_B$ means? Does that mean the answer is yes? – 2017-01-24
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    I'm using $f|_A$ to represent the restriction of $f$ to $A$. But as long as $A$ and $B$ are disjoint, then the answer should be yes. – 2017-01-24
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    What are the chances of seeing a proof or pseudo proof? – 2017-01-24
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    From what the question describes, I'm assuming that there is some total order structure on $C$. Then think of the induced order generated by $f|_A$ and $f|_B$. Each of which are ordered and come from elements in $C$. Then since the ranges are disjoint, we can interpolate these elements in $C$ – 2017-01-24
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    I am half expecting to find a contradiction. The contradiction I'm expecting to see, is that if $f$ does totally order $C$ then there must be some $a\in A,b\in B:f(a)=b$ or $f(b)=a$. – 2017-01-24
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    Maybe I'm not sure how you are defining $f$ as ordering $C$. I'm assuming that there is some natural order on $C$ and $f$ is mapping into that order. – 2017-01-24
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    $f$ is the Collatz function from one odd number to the next so it maps $3$ and $13$ to $5$. $C$ is all odd numbers greater than 1. $A$ is the numbers leading to $1$. $B$ is some hypothetical set of numbers which never lead to $1$. – 2017-01-24
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    If your answer is correct it implies that the Collatz function forms a total ordering of all the odd integers. – 2017-01-24
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    It could be the last step I think in a proof I have been working on – 2017-01-24
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    Is the ordering you are thinking of coming from $a$a$ comes from some iteration of $f$ applied to $b$ or from induced order from the natural numbers? Because the proof I have in mind just says that the order comes from the natural numbers – 2017-01-24
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    it comes from iteration of $f$ – 2017-01-24