Set Theory #1
A great little proof that is simple and a nice way to practice your $\LaTeX$.
Demonstrate that this is a false claim: $C \subset A$ and $C \subset B \Leftrightarrow C \subset (A \cup B)$.
First we examine the righthand implication, that is:
$C \subset A$ and $C \subset B \rightarrow C \subset (A \cup B)$.
Let $x \in C $. Then $x \in A $. So, $x \in A$ or $x \in B$. Hence $x \in (A \cup B)$.
Second, we examine the lefthand implication, that is $C \subset A$ and $C \subset B \leftarrow C \subset (A \cup B)$.
This second statement is false as demonstrated by the following simple counterexample:
$A = [1, 2]$, $B =[1, 3]$, and $C =[2, 3]$.
Then, $A\cup B =[1,2,3]$.
Hence, $C \subset A \cup B$, but, $C \not\subset A$, and $C \not\subset B$.