Question 1080521: A and B are any subsets of some universal set X.
proof or counterexample?
A^c = (A ∪ B)^c ∪ (B\A)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! The claim is true. Look below to see the proof of why the claim is true.
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Let k be any element in set A^c. This means that k is NOT in set A. The value k is anywhere but set A. It could be in set B or somewhere outside both sets A and B.
The goal for this subpart is to show that k is an element of the set on the right side of the equal sign. It doesn't matter which part you pick, (A U B)^c or (B\A), the element k will not be found in set A. The notation B\A means "start with set B and kick out anything found in set A", so clearly k is also a part of this set. Or we can say k is a part of (A U B)^c since this also excludes set A as well. No matter which choice you pick, you'll be outside of set A.
So this means if k is an element of A^c, then k is also an element of either (A U B)^c or (B\A) which is the same as saying k is an element of (A U B)^c U (B\A).
Furthermore, this means A^c is a subset of (A U B)^c U (B\A)
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Let's go in reverse
Let m be some element in (A U B)^c U (B\A). If m is in that set, then either m is in (A U B)^c or (B\A)
Let's assume m is in (A U B)^c. If that's true, then m is NOT in set A because of the complement notation. In this specific case, m is not in set B either
OR we can assume m is an element of (B\A) which means m is in B, but not in A.
So whichever part of (A U B)^c U (B\A) you pick, the idea is that this element is NOT in set A, which is exactly what A^c is saying
So if m is an element of (A U B)^c U (B\A), then m is an element of A^c
In translation, (A U B)^c U (B\A) is a subset of A^c
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In summary, we have shown that A^c is a subset of (A U B)^c U (B\A), which was the first subpart of this proof. We've also shown that (A U B)^c U (B\A) is a subset of A^c, which was the second subpart of the proof. Tying those subparts together and that means A^c = (A U B)^c U (B\A)
This concludes the proof.
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If you prefer a visual approach, then consider this drawing
The rectangle represents the universal set X. This is the entire set of values we care about. It doesn't matter if they are numbers, letters, people, or whatever. The same theories will hold true regardless of the element type.
Inside this universe is the set A and B which overlap somehow. Often it's done with a Venn Diagram. The region that is in A only, and not in B, is represented by the label "A" in the drawing. Likewise, the label "B" means we're in set B but not in set A. The label "C" denotes the overlap between sets A and B. Based on the drawing,  using set notation. Finally, the "D" label represents the region that is NOT in set A and NOT in set B. It is in neither region and outside both.
Hopefully you would agree that ^c) . If not, then consider the set A U B which is the combination of the regions A,B,C. Use a highlighter to shade in all three regions. The region not shaded is the empty space around the circles, which is exactly what set D is. So that's why D is (A U B)^c. In other words "D is everything not in A nor B nor both".
Keep this region D in mind because it will be used shortly.
Moving on, we will focus on B\A now. Go back to a unshaded version of the drawing above. Shade set B but do NOT shade any parts that overlap with A. We only want the stuff in B that doesn't have A in it. That region on the drawing is marked with "B".
The two regions D and B will combine to form basically everything but the regions A and C as shown in this drawing below (light blue is B\A and green is (A U B)^c)
note: the green and blue regions do not overlap
Union these shaded regions together and you get everything but set A. This visually shows that (A U B)^c U (B\A) is the same as A^c, hence showing that A^c = (A U B)^c U (B\A) is true for all elements in A and B.
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