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Chapter 1: Problem 9

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

Short Answer

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

Step by step solution

01

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 \( B \).
02

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 \( C \).
03

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.
04

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 \( x \in B \).
05

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 \( A \cup C \subseteq B \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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 \( B \).

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Most popular questions from this chapter

If \(\boldsymbol{b}>\mathbf{0}\) and \(\boldsymbol{B}>\mathbf{0}\) prove that $$ \frac{a}{b}<\frac{A}{B} \Rightarrow \frac{a}{b}<\frac{a+A}{b+B}<\frac{A}{B} $$ Further, if \(\boldsymbol{p}\) and \(\boldsymbol{q}\) are positive integers such that $$ \frac{7}{10}<\frac{p}{q}<\frac{11}{15} $$ what is the least value of \(\boldsymbol{q}\) ?

On \(\mathbb{Q} \cap]-1 ; 1[\) define the binary operation \(\otimes\) by $$ a \otimes b=\frac{a+b}{1+a b}, $$ where juxtaposition means ordinary multiplication and \(+\) is the ordinary addition of real numbers. Prove that \((\mathrm{Q} \cap]-1 ; 1[, \otimes)\) is an abelian group by following these steps. 1\. Prove that \(\otimes\) is a closed binary operation on \(\mathrm{Qn}]-1 ; 1[.\) 2\. Prove that \(\otimes\) is both commutative and associative. 3\. Find an element \(e \in \mathbb{Q} \cap]-1 ; 1[\) such that \((\forall a \in Q \cap]-1 ; 10\) (e\otimes \(a=a)\) 4\. Given e as above and an arbitrary element \(a \in \mathrm{Q} \cap]-1 ; 1[\), solve the equation \(\boldsymbol{a} \otimes \boldsymbol{b}=e\) for \(\boldsymbol{b}\).

(Putnam, 1972) Let \(\mathscr{S}\) be a set and let * be a binary operation of \(\mathscr{F}\) satisfying the laws \(\forall(x, y) \in \mathscr{S}^{2}\) $$ \begin{aligned} &x *(x * y)=y \\ &(y+x) * x=y . \end{aligned} $$ Shew that \(*\) is commutative, but not necessarily associative.

(Putnam 1948) If \(n\) is a positive integer, demonstrate that $$ \lfloor\sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4 n+2} \rrbracket $$

Let \(n\) be a strictly positive integer, let \(a_{k} \geq 0,1 \leq\) \(k \leq n\) be real numbers such that \(a_{1} \geq a_{2} \geq \cdots \geq a_{n}\), and let \(b_{k}, 1 \leq\) \(k \leq n\) be real numbers. Assume that for all indices \(\boldsymbol{k} \in\\{\mathbf{1 , 2}, \ldots, \boldsymbol{n}\\}\), $$ \sum_{i=1}^{k} a_{i} \leq \sum_{i=1}^{k} b_{i} $$ Prove that $$ \sum_{i=1}^{n} a_{i}^{2} \leq \sum_{i=1}^{n} b_{i}^{2} $$
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