Math
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:
Backlinks (2)
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Knowledge is a paradox. The more one understand, the more one realises the vastness of his ignorance.