Solutions to Kolmogorov and Fomin's Introductory Real Analysis

Sun, 20 Jul 2025 02:42:32 +1100

Set Theory

Sets and Functions

Problem

Prove that if \(A\cup B = A\) and \(A \cap B = A\), then \(A = B\).

Solution

To show \(A=B\), show \(A\subseteq B\) and \(B\subseteq A\). Suppose \(x\in B\), then we know by definition that \(x\in (A\cup B)\) if \(x\in B\) or \(x\in A\). Which then implies that \(x\in A\) from rule 1. Thus \(B\subseteq A\). Now suppose \(x\in A\) which implies \(x\in (A\cap B)\) (rule 2). The definition of this means that \(x\in A\) and \(x\in B\). \(\therefore x\in A \implies x\in B\), i.e. \(A\subseteq B\), so \(A=B\).

Problem

Show that in general \((A-B)\cup B \neq A\).

Solution

This only holds for \(B\subseteq A\). We proceed by counterexample. Let \(A={1,2}, B={3,4}\). Then \((A-B) = {1,2}\) and \((A-B)\cup B = {1,2,3,4} \neq {1,2}\).

Problem

Let \(A = {2,4,…,2n,…}\) and \(B = {3,6,…,3n,…}\). Find \(A\cap B\) and \(A-B\)

Solution

\[ A\cap B = {6n \mid n\in \mathbb{N}} \] \[ A - B = {2n \mid n\in \mathbb{N}, 2n \not\in {6m \mid m\in \mathbb{N}}} \]

Problem

Prove that:

()

\((A-B)\cap C = (A\cap C) - (B\cap C)\)

Solution:

Let \(x\in (A-B)\cap C\). Then:

Kolmogorov on Aayush Bajaj's Augmenting Infrastructure

Solutions to Kolmogorov and Fomin's Introductory Real Analysis

Set Theory#

Sets and Functions#

Set Theory

Sets and Functions