Sample space and total possible events Taken from my book:
How many events are there associated with the roll of one die? Solution: Each event corresponds to a subset of {1,2,3,4,5,6}. there are $2^6$ subsets, so there are $2^6$ possible events.
Now my question is: why $2^6$ events? If you roll only one dice shouldn't the possible events be only 6? {1}, {2}, {3}, {4}, {5} and {6}?
 A: 2^6 events that is all subsets are not events of roll of one die . 
2^6also includes,
{1,2} ; {1,3} ; {1,2,3,4,} ; {4,5,6}. Sunder is completely different than event.
A: Can it happen that after rolling the die the upward face of it will show an even number? Yes, and this motivates to accept the set $\{2,4,6\}$ as an event. We can go further in this. Let $S$ be a subset of $\{1,2,3,4,5,6\}$. Then can it happen that the number on the upward face of the die will be an element of $S$? Yes, so... Wait a minute. Not if $S$ is empty of course. Nevertheless also $\varnothing$ is accepted as an event. The fact that it corresponds with something that cannot happen does not cause any troubles if we give this event probability $0$, as we always do in probability.
A: You have the terms "event" and "outcome" confused.
The outcomes of the sample space are: $1, 2, 3, 4, 5,$ and $6$.   These are the elements of the sample space:   $\{1, 2, 3, 4, 5,6\}$ .
An event is a set of outcomes.   Such as the event of rolling an even number: $\{2,4,6\}$, the event of rolling a number greater than four: $\{5,6\}$, and such.
There are $2^6$ subsets of the sample space, which are all the plausible events: $\{\}, \{1\}, \{2\}, \{3\}, \{4\}, \{5\}, \{6\}, \{1,2\}, \ldots, \{5,6\}, \{1,2,3\},\ldots,\ldots,\{1,2,3,4,5,6\}$.
The set of these form a sigma-algebra, but that's another topic.
