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Do you know how to solve this problem? I have two sets and need to solve $A \cup X = B. $

Thanks a lot for your help.

Philippe Malot
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dzoni
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  • I guess the two sets you have are $A$ and $B$, and you need to find $X$, right? – Philippe Malot Oct 20 '13 at 13:36
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    Could explain exactly what the problem is and what have you tried? – user77404 Oct 20 '13 at 13:39
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    "Do you know how to solve this problem?" Yes: by drawing a Venn diagram. – Did Oct 20 '13 at 13:40
  • @Did As I say to my second years every week: A Venn diagram is not a proof! – user1729 Oct 20 '13 at 13:48
  • (Also, @dzoni how on earth can this problem be solved? Do you have explicit examples of $A$ and $B$? or are you wanting to write $X$ in terms of $A$ and $B$?) – user1729 Oct 20 '13 at 13:48
  • i don't have any explicit examples of A and B. I have only a task:" You have two sets A and B. Solve $A \cup X =B$ and that's all. I have tried do it by venna diagram, but i lost in it and still don't have a result. – dzoni Oct 20 '13 at 13:55
  • @user1729 A Venn diagram is not a proof and a Venn diagram is definitely a short and easy way to guess the answer and to be ready to provide a proof. (If the message of your comment is that one should stay away from Venn diagrams, then I squarely disagree with the message.) – Did Oct 20 '13 at 14:06
  • @Did No no, I tell them that a Venn diagram is not a proof but helps you to find the proof. – user1729 Oct 20 '13 at 14:15

2 Answers2

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Note that if in fact there exists a set $B = A \cup X$, then $A\subseteq B$ AND $X\subseteq B$.

If $A$ and $X$ are disjoint, so that $A \cap X = \varnothing,$ then $$A\cup X = B \iff (A\cup X)\setminus A = B\setminus A \iff X = B\setminus A$$

If they are not disjoint, then you we need to know what $A \cap X$ is, since we'd have to look at $$X = (B\setminus A) \cup(A \cap X)$$ where $$B\setminus A, \;\text{ means }\;(B \;\text{ set-minus }\; A).$$

amWhy
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  • Looks like you were busy too! +1 – Amzoti Oct 21 '13 at 00:30
  • Oh, nice work on the talk! Finishing the first pass is the hardest part...the revision and fine-tuning usually "fall out" after that. (At least that's true for me - Once I draft an outline, better yet, a "rough draft", I'm in good shape - it's getting past the "blank page" that trips me up!) – amWhy Oct 21 '13 at 00:38
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If $A\cup X=B$ has some solution $X$ then $A\subseteq B$.

If $A\subseteq B$ then $A\cup X=B$ if and only if there exists $C\subseteq A$ such that $X=C\cup(B\setminus A)$.

Did
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