0 votes 0 votes $\text{Theorem}:$ Let $S$ be an infinite countable set. Then its powerset $2^S$ is not countable. Why does the argument in Theorem fail when $S$ is finite$?$ Theory of Computation peter-linz peter-linz-edition5 theory-of-computation proof turing-machine recursive-and-recursively-enumerable-languages + – Rishi yadav asked Mar 16, 2019 Rishi yadav 203 views answer comment Share Follow See 1 comment See all 1 1 comment reply prashant jha 1 commented Mar 17, 2019 reply Follow Share $2^{S}$ is simply the power set of set $S$, that is just the set of all possible subsets , when $S$ is finite there's always a limit of the size of the power set which is denoted by $2^{|S|}$ , when $S$ is finite then $|S|$ is also finite (which means $|S|$ can be corresponded with a Natural Number) and $2^{|S|}$ when $|S|$ is finite is always countable. 0 votes 0 votes Please log in or register to add a comment.
