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Assume ● $A,B,C$ are nonempty sets and
$A\cup C$ and $B\cup C$ are numerically equivalent and
$A\cap C=B\cap C=\emptyset$.

Prove or disprove that $A,B$ are numerically equivalent.

Intuitively, for a finite set $A,B$, the statement should be true. But when considering infinite sets, then I don't know how to prove or disprove it.

Edited: "Denumerable" has been changed to "numerically equivalent."

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    I have never seen the term "numerically equivalent". What does it mean?2012-11-03

2 Answers 2

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By hypothesis there is a bijection $h:A\cup C\to B\cup C$, $A\cap C=B\cap C=\varnothing$, and you want to show that there is a bijection from $A$ to $B$. If you were lucky enough to find that $h[C]=C$, you’d be done, because then $h[A]$ would have to be $B$, and $h\upharpoonright A$, the restriction of $h$ to $A$, would be a bijection from $A$ to $B$. Unfortunately, there’s no reason to hope that $h[C]=C$. If $C$ is infinite, $h[C]$ might be a proper subset of $C$, and then we could easily have $|A|>|B|$.

In fact, if $C$ is infinite the result is false; here’s a counterexample. Let $C=\Bbb Z^+$, the set of positive integers, $A=\{0\}$, and $B=\Bbb Z\setminus\Bbb Z^+$, the set consisting of $0$ and the negative integers. Then $A\cup C=\Bbb N$, $B\cup C=\Bbb Z$, and

$h:\Bbb N\to\Bbb Z:n\mapsto\begin{cases} \frac{n}2,&\text{if }n\text{ is even}\\ -\frac{n+1}2,&\text{if }n\text{ is odd} \end{cases}$

is a bijection from $A\cup C$ to $B\cup C$. Obviously, however, $A$ and $B$ do not have the same cardinality.

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If by numerically equivalent you mean the same cardinality, all you can conclude for infinite $C$ is that $|A| \le |C|$ and the same for $B$ or that $|A|=|B|$ (in which case they can be larger than $C$).

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    @MartinSleziak: you are right. I read the infinite case to imply that $C$ was infinite, but even there they could be larger if they are equal cardinality. I'll add it.2012-11-03