# Show that the collection of all subsets of a finite sample space satisfies the 3 conditions to be called a collection of events.

As title. Suppose that the sample space $S$ of some experiment is finite. Show that the collection of all subsets of $S$ satisfies the three conditions required to be called the collection of events.

So sample space $S$ is finite, with $N$ possible outcomes, where $N$ is a positive integer. Thus $S$ has the form $S = \{w_1, ..., w_N\}$, where $w_i$ represents the $i$-th possible outcome, for $i \in \{1, ..., N\}$.

This sample space contains $2^N$ possible events like a power set.

We can construct a probability measure by defining probabilities $p_i$ for each particular outcome $w_i \in S$.

$P[w_i] = p_i, \forall i \in \{1, ..., N\}$

I was thinking of defining the probability measure to all events $A \subseteq S$ by defining $P(\emptyset)=0$, and $P(A)$ for each non-empty subset $A \subseteq S$ as follows:

$P(A) = \sum_{w_i \in A} P(w_i)$

where the summation on the right-hand side represents a sum over each of the elements $w_i$ in the set $A$.

My question is, am I on the right track, if so, how do I show that this definition satisfies the three axioms of probability? That is,

1. $S$ must be an event.

2. If $a \in S$ is an event, then so is $a^c$.

3. If $a_1, ..., a_n$ is a countable collection of events, then $\bigcup_{i=1}^{\infty} a_i$ is also an event.

• It is not necessary to assign probabilities etc. for the task you have to do. You need to only verify that the collection of all subsets of the sample space (there are $2^n$ such subsets) includes the sample space, complements of all subsets, and countable unions of subsets. So start! Can $S$ be regarded as a subset of itself and thus is included in the collection of all subsets of $S$? If $A$ is a subset of $S$, is $A^c = S-A$ a subset of $S$ also? and so on. – Dilip Sarwate Sep 2 '14 at 1:57

For example, is $S$ in our collection? Sure, every subset of $S$ is in our collection, and $S$ is a subset of itself.
• I think I'm confused with the difference of $S$ and the collection of all subsets of $S$. Say if $S = \{1,2\}$, then the collection of all subsets of $S$ will be $\{\emptyset\}, \{1\}, \{2\}, \{1,2\}$ right? If I let $A = \{1\}$, then what does $A^c$ look like? $A^c=\{2\}$ or $A^c=\{\emptyset\}, \{2\}, \{1,2\}$? @Dilip Sarwate as well. Thanks you guys in advance! – PandaMan Sep 2 '14 at 2:13
• If $A=\{1\}$ then $A^c=\{2\}$. We want the set of all elements of $S$ that are not in $A$. – André Nicolas Sep 2 '14 at 2:16