Analysis
2026-01-19
Basic Topology
Numerical Sequences and Series
Continuity
Differentiation
The Riemann-Stieltjes Integral
Sequences and Series of Functions
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:
This front matter follows the pattern of other files in the blog directory, includes math support for the calculations, and reflects the content about birthday prize money analysis with mathematical calculations.
| Years | Prize | Recipient |
|---|---|---|
| 21 | 50,20 | Aarav, Jisu |
| 22 | 50,100 | Erick, Sarro |
| 23 | 100,150 | Unclaimed, Aarav |
| 24 | 300 | Unclaimed |
rule: take the most expensive offering and then double it: 21: 50,20 | 50 22: 50, 100 | 100 23: 100, 150 |