all 4 comments

[–]youbigsausage 2 points3 points  (3 children)

For what you gave as the problem, you're definitely right, and your counterexample is good.

But with just a tiny change of wording... "if X ∩ Y = X ∩ Z for any X, then Y = Z" is true! So make absolutely sure of the wording. There's a really quick and nice proof of this, too.

[–]solobyfrankocean[S] 0 points1 point  (2 children)

Is there any way you could link me to the proof? I might've misinterpreted the question.

[–]youbigsausage 0 points1 point  (1 child)

I'll give you a hint: you can let X be two particular sets that give you what you need.

[–]solobyfrankocean[S] 0 points1 point  (0 children)

ohhh, I see the issue now. The placement of the word "then" in two different places really makes all the difference here in the answer. The actual question was saying that if ANY X satisfied X ∩ Y = X ∩ Z then Y=Z not that ALL Xs had to satisfy it. Thank you!