# Event Space/Examples/Arbitrary Event Space on 6-Sided Die

## Example of Event Space

Let $\EE$ be the experiment of throwing a standard $6$-sided die.

The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.

Let $\FF$ be the arbitrary set of subsets of $\Omega$ defined as:

$\FF = \set {\O, \set {1, 2}, \set {3, 4}, \set {5, 6}, \set {1, 2, 3, 4}, \set {3, 4, 5, 6}, \set {1, 2, 5, 6}, \Omega}$

Then $E$ is an event space of $\EE$.

## Proof

Event Space Axiom $(\text {ES} 1)$

It is specified that $\O \in \FF$.

Thus axiom $(\text {ES} 1)$ is fulfilled.

$\Box$

Event Space Axiom $(\text {ES} 2)$

It is specified that $\Omega \in \FF$.

Thus axiom $(\text {ES} 2)$ is fulfilled.

$\Box$

Event Space Axiom $(\text {ES} 3)$

We investigate the unions of elements of $\Omega$.

From Union with Superset is Superset, it is unnecessary to investigate the union of any element of $\Omega$ with a subset of it.

We continue:

 $\ds \set {1, 2} \cup \set {3, 4}$ $=$ $\ds \set {1, 2, 3, 4}$ $\ds \in \Omega$ $\ds \set {1, 2} \cup \set {5, 6}$ $=$ $\ds \set {1, 2, 5, 6}$ $\ds \in \Omega$ $\ds \set {3, 4} \cup \set {5, 6}$ $=$ $\ds \set {3, 4, 5, 6}$ $\ds \in \Omega$ $\ds \set {1, 2} \cup \set {3, 4} \cup \set {5, 6}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$ $\ds \set {1, 2, 3, 4} \cup \set {1, 2, 5, 6}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$ $\ds \set {1, 2, 3, 4} \cup \set {3, 4, 5, 6}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$ $\ds \set {1, 2, 5, 6} \cup \set {3, 4, 5, 6}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$ $\ds \set {1, 2} \cup \set {3, 4, 5, 6}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$ $\ds \set {3, 4} \cup \set {1, 2, 5, 6}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$ $\ds \set {5, 6} \cup \set {1, 2, 3, 4}$ $=$ $\ds \set {1, 2, 3, 4, 5, 6} = \Omega$ $\ds \in \Omega$

$\Box$

All the event space axioms are seen to be fulfilled by $\powerset \Omega$.

Hence the result.

$\blacksquare$