Set Theory #1

Thu 10 April 2014

Filed under math

Tags math

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:

Choose:

$A = [1, 2]$, $B =[1, 3]$, and $C =[2, 3]$.

Then, $A\cup B =[1,2,3]$.
Hence, $C \subset A \cup B$, but, $C \not\subset A$, and $C \not\subset B$.


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