Solved VENN DIAGRAMS, ALGEBRA OF SETS, DUALITY 2.70. The | Chegg.com - Free Printable

Let's solve each of the problems step by step.

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Problem 2.70: Shade the following sets using the Venn diagram

We are given a Venn diagram with three overlapping sets: $ A $, $ B $, and $ C $. We need to shade the regions corresponding to:

#### (a) $ A \setminus (B \cup C) $

This is the set of elements that are in $ A $ but not in $ B \cup C $.

- First, $ B \cup C $: all elements in $ B $ or $ C $ or both.
- Then $ A \setminus (B \cup C) $: parts of $ A $ that do not overlap with $ B $ or $ C $.

So we shade the region only in $ A $, excluding any part where $ A $ overlaps with $ B $ or $ C $.

✔ Shade: The part of circle $ A $ that does not intersect with $ B $ or $ C $ — this is the region exclusive to $ A $.

> This is the area inside $ A $ but outside both $ B $ and $ C $.

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#### (b) $ A^c \cap (B \cap C) $

Break it down:
- $ A^c $: complement of $ A $ → everything outside $ A $
- $ B \cap C $: elements in both $ B $ and $ C $
- Intersection: elements that are in $ B \cap C $ and not in $ A $

So we want the region where $ B $ and $ C $ overlap, but not in $ A $

✔ Shade: The intersection of $ B $ and $ C $ that lies outside $ A $

> That is, the lens-shaped region where $ B $ and $ C $ overlap, but not overlapping with $ A $

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#### (c) $ (A \cup C) \cap (B \cup C) $

Use distributive law or analyze directly:

We can simplify:
$$
(A \cup C) \cap (B \cup C) = (A \cap B) \cup C
$$
(using distributive law: $ (X \cup C) \cap (Y \cup C) = (X \cap Y) \cup C $)

But let’s interpret geometrically:

- $ A \cup C $: all of $ A $ and $ C $
- $ B \cup C $: all of $ B $ and $ C $
- Their intersection: elements that are in both $ A \cup C $ and $ B \cup C $

So:
- Any element in $ C $ is in both → so entire $ C $ is included
- Also, elements in $ A $ and $ B $ (but not necessarily $ C $) are included if they're in both unions

Wait: for an element to be in both $ A \cup C $ and $ B \cup C $, it must be:
- In $ A $ or $ C $, and
- In $ B $ or $ C $

So:
- If it's in $ C $: automatically in both → include
- If not in $ C $, then must be in both $ A $ and $ B $

So overall:
$$
(A \cup C) \cap (B \cup C) = (A \cap B) \cup C
$$

✔ Shade: All of $ C $, plus the region where $ A $ and $ B $ overlap (but not in $ C $)

> So: the entire circle $ C $, and the part of $ A \cap B $ that is outside $ C $

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✔ Summary for 2.70:

| Part | Set | Region to Shade |
|------|-----|------------------|
| (a) | $ A \setminus (B \cup C) $ | Only $ A $, not overlapping $ B $ or $ C $ |
| (b) | $ A^c \cap (B \cap C) $ | $ B \cap C $ but not in $ A $ |
| (c) | $ (A \cup C) \cap (B \cup C) $ | Entire $ C $, plus $ A \cap B $ outside $ C $ |

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Problem 2.71: Write the dual of each equation

In set theory, duality means replacing:
- $ \cup $ with $ \cap $
- $ \cap $ with $ \cup $
- $ U $ (universal set) with $ \emptyset $ (empty set)
- $ \emptyset $ with $ U $
- Complements remain as they are (since $ A^c $ is self-dual in this context)

Also, identities like $ A \cup A = A $ have duals $ A \cap A = A $

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#### (a) $ A \cup (A \cap B) = A $

Dual:
- Replace $ \cup $ with $ \cap $
- Replace $ \cap $ with $ \cup $
- So: $ A \cap (A \cup B) = A $

✔ Dual: $ A \cap (A \cup B) = A $

This is known as the absorption law for intersection.

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#### (b) $ (A \cap B) \cup (A^c \cap B) \cup (A \cap B^c) \cup (A^c \cap B^c) = U $

Let’s rewrite:
- This is the union of all possible combinations of $ A $ and $ B $: essentially, it covers the whole universal set $ U $
- It includes all four regions in a two-set Venn diagram.

Now apply duality:
- Replace $ \cup $ with $ \cap $
- Replace $ \cap $ with $ \cup $
- Replace $ U $ with $ \emptyset $

So:
- Left side becomes:
$$
(A \cup B) \cap (A^c \cup B) \cap (A \cup B^c) \cap (A^c \cup B^c)
$$
- Right side becomes $ \emptyset $

But wait: is this correct?

Let’s check logic.

Original: Union of all 4 basic regions → equals $ U $

Dual: Intersection of the complements of those regions? Not quite.

Actually, better way: Take the dual of the expression by swapping operations and constants.

So:
- $ (A \cap B) \to (A \cup B) $
- $ (A^c \cap B) \to (A^c \cup B) $
- $ (A \cap B^c) \to (A \cup B^c) $
- $ (A^c \cap B^c) \to (A^c \cup B^c) $
- $ \cup $ → $ \cap $
- $ = U $ → $ = \emptyset $

So dual is:
$$
(A \cup B) \cap (A^c \cup B) \cap (A \cup B^c) \cap (A^c \cup B^c) = \emptyset
$$

Is this true?

Let’s test: What is the intersection of these?

Note:
- $ A \cup B $: everything except $ A^c \cap B^c $
- $ A^c \cup B $: everything except $ A \cap B^c $
- $ A \cup B^c $: everything except $ A^c \cap B $
- $ A^c \cup B^c $: everything except $ A \cap B $

So their intersection is only the region that is in all four.

But no point can be in all four unless it's in every region — but actually, think about it:

The only region missing from $ A \cup B $ is $ A^c \cap B^c $

Similarly, missing from others.

But the intersection of all four expressions is the set of elements that satisfy all conditions.

But notice:
- $ (A \cup B) \cap (A^c \cup B) = B $ (by distribution)
- Then $ B \cap (A \cup B^c) = B \cap A $ since $ B \cap B^c = \emptyset $
- Then $ A \cap B \cap (A^c \cup B^c) $

But $ A \cap B \subseteq A $, so $ A \cap B \cap (A^c \cup B^c) $: elements in $ A \cap B $ and either $ A^c $ or $ B^c $

But $ A \cap B \subseteq A $, so $ A \cap B \cap A^c = \emptyset $, and $ A \cap B \cap B^c = \emptyset $

So entire thing is empty.

Thus, yes, the dual expression equals $ \emptyset $

✔ Dual:
$$
(A \cup B) \cap (A^c \cup B) \cap (A \cup B^c) \cap (A^c \cup B^c) = \emptyset
$$

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Problem 2.72: Use laws to prove $ (A \cap B) \cup (A \cap B^c) = A $

We will use basic set laws such as:
- Distributive Law
- Complement Law
- Identity Law
- etc.

Start with left-hand side:
$$
(A \cap B) \cup (A \cap B^c)
$$

Apply Distributive Law: $ X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z) $, but here it's reverse.

Factor out $ A $:

$$
= A \cap (B \cup B^c)
$$

Because $ (A \cap B) \cup (A \cap B^c) = A \cap (B \cup B^c) $ — this is distributive law applied backwards.

Now:
- $ B \cup B^c = U $ (complement law)

So:
$$
= A \cap U
$$

And:
- $ A \cap U = A $ (identity law)

Therefore:
$$
(A \cap B) \cup (A \cap B^c) = A
$$

✔ Proved.

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✔ Final Answers:

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#### 2.70: Shading Instructions

(a) $ A \setminus (B \cup C) $: Shade only the part of $ A $ that does not intersect $ B $ or $ C $

(b) $ A^c \cap (B \cap C) $: Shade the region where $ B $ and $ C $ overlap, but not in $ A $

(c) $ (A \cup C) \cap (B \cup C) $: Shade all of $ C $ and the region where $ A $ and $ B $ overlap outside $ C $

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#### 2.71: Duals

(a) $ A \cup (A \cap B) = A $ → Dual: $ A \cap (A \cup B) = A $

(b) $ (A \cap B) \cup (A^c \cap B) \cup (A \cap B^c) \cup (A^c \cap B^c) = U $
→ Dual: $ (A \cup B) \cap (A^c \cup B) \cap (A \cup B^c) \cap (A^c \cup B^c) = \emptyset $

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#### 2.72: Proof

$$
(A \cap B) \cup (A \cap B^c) = A \cap (B \cup B^c) = A \cap U = A
$$

✔ Proven using distributive law, complement law, and identity law.

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