Is it true that |A ∩ D| < |B ∩ D| if only if |A| < |B|?

Question

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?

Answer 1

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 $B$.