As title says, if a set $\Sigma$ of alphabets is of cardinality $k$, does $\Sigma^n$ have cardinality of $k^n$? This seems to be the case because for each character of the string of length $n$, you have $k$ choices, so $k^n$. Is this right?
Asked
Active
Viewed 34 times
0
-
Yes i think yo say correct – erfan soheil Jan 29 '15 at 11:24
-
Yes. That's what the notation is suggesting. – Qiaochu Yuan Jan 30 '15 at 09:46
1 Answers
0
For any set $S$ of cardinality $k$, $S^n$ has cardinality $k^n$. This is fundamental property of Cartesian product
Xoff
- 10,310