I'm having problems answering the following question:

Give a counter example to show that the following statement is false:

For any sets $A$ and $B$, if $|A \cap B| < |A|$ then $|A|>|B|$

Presumably we can't be sure whether $|A|>|B|$ as we don't know how many elements there are in $B$ which aren't shared with $A.$


You're on the right track. See what happens when $A$ and $B$ have no elements in common at all. Can you build a counterexample in this case?

  • $\begingroup$ Well let's assume that |A| = 5 and |B| = 9| And |A Intersect B| = 0, then |A| is not > |B|. $\endgroup$ – M-R Oct 11 '15 at 22:11
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    $\begingroup$ Yep, this works just fine. $\endgroup$ – AnatolyVorobey Oct 11 '15 at 22:13
  • $\begingroup$ I've provided a fairly specific example. Is there a more abstract counterexample I could provide? $\endgroup$ – M-R Oct 11 '15 at 22:15
  • $\begingroup$ @MartinRand Your example is fine, but it would perhaps be more elegant to use a smaller example, like two disjoint one-element sets. $\endgroup$ – Andreas Blass Oct 11 '15 at 22:19
  • $\begingroup$ @MartinRand Take any three mutually exclusive (aka disjoint) sets $X, Y, Z$ where $0 < \lvert Y\rvert \leq \lvert Z\rvert$ . Call $A=X\cup Y$ and $B=X\cup Z$. Done. $\endgroup$ – Graham Kemp Oct 12 '15 at 0:02

You have the right idea. So you should be able to construct your counterexample if you start with some sets that already satisfy $|A\cap B|<|A|$, and then add enough fresh elements to $B$ to make $|A|\not>|B|$.


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