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}?
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.
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.
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.