Is it true that $|A \cap D| < |B \cap D|$ if only if $|A| < |B|$? For finite sets, at least, I thought that there are some sets, for example, if you have $A=\{1,2,3\}$, $B=\{4,6,8,9\}$, and $D=\{2,3,6\}$ that means $(A \cap D)=\{2,3\}$ and $(B \cap D)=\{6\}$ and that means $|A \cap D| < |B \cap D|$. But, how you prove it if $A$, $B$, and $D$ are infinite sets?
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$\begingroup$ Consider $A=D=\{1\},B=\{2,3\}$. $|A|<|B|$ but $|A\cap D|>|B\cap D|$. So your claim is false even for finite sets. $\endgroup$– Kyan CheungCommented Jun 17, 2021 at 8:25
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$\begingroup$ Doesn't $A = \mathbb Z, B = D = \mathbb Q$ work just fine? (as a counterexample I mean) $\endgroup$– Stephen DonovanCommented Jun 17, 2021 at 8:26
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2$\begingroup$ It isn't true, in general, for either finite or infinite sets. One of the critical questions is whether there is some element in $D$ that is in $B$ but is not in $A$. Just because $A$ has fewer elements than $B$ does not imply that one of the elements that $B$ contains that is missing in $A$ also happens to be in $D$. Another issue is that just because $A$ has fewer elements than $B$, it could still have more elements in common with $D$ than $B$ does. $\endgroup$– user2661923Commented Jun 17, 2021 at 8:27
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2$\begingroup$ @StephenDonovan $|\mathbb Z|=|\mathbb Q|$ (see math.stackexchange.com/questions/1311/…) so your counterexample does not work. $\endgroup$– Kyan CheungCommented Jun 17, 2021 at 8:29
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$\begingroup$ @Kyky Sorry I goofed, I don't know why I didn't see that the integers were also rational numbers. (I may need sleep) Consider instead then $A = \{z \in \mathbb C : \text{Re}(z) = 0\},$ $B = D = \mathbb R.$ $\endgroup$– Stephen DonovanCommented Jun 17, 2021 at 8:42
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1 Answer
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The statement is false in both direction of the "if and only if", and both for finite and for infinite sets.
Finite example: Take $A=\{1,2\}$, $B=\{3\}$ and $D=\{3\}$.
$A\cap D=\varnothing$, while $B\cap D=\{3\}$.
Infinite example:
\begin{align}
A&=\{(x,y)\in\Bbb R^2\mid x=0\}\\
B&=\{(x,y)\in \Bbb R^2 \mid x=1\text{ and }y\text{ is a natural number}\}\\
D&=\{(x,y)\in\Bbb R^2\mid x=1\}
\end{align}
Then $|A|=|D|=|\Bbb R|=2^{\aleph_0}$ is uncountable, while $|B|=|\Bbb N|=\aleph_0$ is countable.
$A\cap D=\varnothing$, while $B\cap D=B$.
For the other direction of the "if and only if" you just switch the roles of $A$ and $B$.
