Problem 9

Question

Prove that $A \subseteq B$ and $\quad C \subseteq B \Longrightarrow A \cup C \subseteq B .$

Step-by-Step Solution

Verified

Answer

Since elements from either A or C are in B, their union is also within B.

1Step 1: Understand What Needs to be Proven

We need to prove that if $ A \subseteq B $ and $ C \subseteq B $, then $ A \cup C \subseteq B $. This means every element of $ A \cup C $ should also be an element of $ B $.

2Step 2: Definition Review

Recall that $ A \subseteq B $ means every element of $ A $ is also an element of $ B $, and $ C \subseteq B $ means every element of $ C $ is also an element of $ B $. $ A \cup C $ is the set containing all elements that are in either $ A $ or $ C $.

3Step 3: Element Membership in Union

Take an arbitrary element $ x $ in $ A \cup C $. By definition of union, $ x $ must be in either $ A $ or $ C $ or in both.

4Step 4: Apply Subset Condition

Since $ A \subseteq B $, if $ x \in A $, then $ x \in B $. Similarly, since $ C \subseteq B $, if $ x \in C $, then $ x \in B $.

5Step 5: Conclude Entitlement to Union Belonging B

Since $ x \in A $ implies $ x \in B $ and $ x \in C $ also implies $ x \in B $, every element $ x $ in $ A \cup C $ must be in $ B $, thus proving that $ A \cup C \subseteq B $.

Key Concepts

SubsetUnion of SetsElement Membership in Sets

Subset

In set theory, the concept of a subset is fundamental. A set $ A $ is considered a subset of another set $ B $ when every element in $ A $ is also an element in $ B $. This relationship is expressed as $ A \subseteq B $. It’s like saying that $ A $ fits perfectly into $ B $, without any element in $ A $ being left out of $ B $. Therefore, knowing an element is part of $ A $ is enough to assure us it is also a part of $ B $.

Example: Let $ A = \{ 1, 2 \} $ and $ B = \{ 1, 2, 3 \} $, here $ A \subseteq B $ because both elements 1 and 2 are inside $ B $.
If even one element of $ A $ is not in $ B $, then $ A $ is not a subset of $ B $.

Union of Sets

Union of sets is another key idea. When finding the union of two sets $ A $ and $ C $, represented by $ A \cup C $, you are essentially creating a new set. This new set consists of any and all distinct elements that are part of either $ A $, $ C $, or both. It is like pooling elements from various groups together, ignoring duplicates.

For instance, if $ A = \{ 1, 3, 5 \} $ and $ C = \{ 2, 3, 4 \} $, then $ A \cup C = \{ 1, 2, 3, 4, 5 \} $.
This broad collection ensures no opportunity of leaving any eligible element out of the combined set.

It's important that, when $ A \subseteq B $ and $ C \subseteq B $, both $ A $ and $ C $ contribute their elements to $ A \cup C $, which then should naturally still lie within the larger set $ B $.

Element Membership in Sets

Element membership is a simple yet significant concept in set theory. It specifies whether an individual element belongs to a specific set. If an element $ x $ is in set $ A $, it is denoted as $ x \in A $. Understanding membership helps in understanding relationships between sets, such as when analyzing subsets or unions.

Think of it like having a box of different objects. If you find a ball in the box, you can say, "the ball is a member of the box's contents."
If we know $ x $ is in $ A $, and $ A \subseteq B $, $ x $ must also be in $ B $.

When discussing the union $ A \cup C $ and ensuring every element still fits in $ B $, recognizing element membership ensures that no rules of set belongingness are violated through the transfer from $ A $ or $ C $ into $ B $.

Problem 9

Problem 10

Other exercises in this chapter

Problem 9

Demonstrate that $$ \sum_{k=0}^{n}(k-n p)^{2}\left(\begin{array}{l} n \\ k \end{array}\right) p^{k}(1-p)^{n-k}=n p(1-p) $$

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Problem 9

Prove that $f: x \rightarrow \frac{x-1}{x+1}$ is a bijection and find $f^{-1} .$

View solution

Problem 10

Shew that $\forall x>0, x$ is farther from $\sqrt{5}$ than $\frac{2 x+5}{x+2}$ is.

View solution

Problem 10

Prove that $\forall(a, b, c) \in \mathbb{R}^{3}$, $$ a^{2}+b^{2}+c^{2} \geq a b+b c+c a $$

View solution