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My friend told me that say you have an uncountable set $A$ and a countable set $B$, then the intersection of these two sets is the empty set. But wouldn't something like $A = [0,1]$ and $B = \{1, 2\}$ have the intersection of $\{1\}$? Also is there a way to prove this? Thanks!

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    $\begingroup$ Friends do say some peculiar things.... $\endgroup$ – Lord Shark the Unknown Oct 10 '17 at 6:04
  • $\begingroup$ Your counterexample is correct. $\endgroup$ – Qudit Oct 10 '17 at 6:05
  • $\begingroup$ With friends like that, you will never need enemies $\endgroup$ – Mark Fischler Oct 10 '17 at 6:05
  • $\begingroup$ Yeah I figured he was wrong so does this also mean that the intersection of countable and uncountable set is always countable? Seems like it should be. $\endgroup$ – aspookyghost20 Oct 10 '17 at 6:07
  • $\begingroup$ Yes, you cannot generate an uncountable set from an intersection (subset) of a countable set. It will be countable or empty. Proof by contradiction $\endgroup$ – Daniel Oct 10 '17 at 6:43
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You have disproved your friend's claim.

$A \cap B \subseteq B$.

Hence $A \cap B$ is countable but as you have shown, it need not be empty.

Another counter example would be let $A = \mathbb{R}$ and $B = \mathbb{Q}$, then $A \cap B = \mathbb{Q} \neq \emptyset.$

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