0 votes 0 votes I didn't get the solution. HeadShot asked Jan 1, 2019 HeadShot 354 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Hemanth_13 commented Jan 1, 2019 reply Follow Share For example if L={$\epsilon$,a,b} then $L-{\epsilon}$ = {a,b} but $L^+ $ = {{$\epsilon$},{a},{b},{a,b}} --> the $\epsilon$ here is as it in the language we placed the it but $L^*$ also generates one $\epsilon$ which is removed in $L^+$. 0 votes 0 votes HeadShot commented Jan 1, 2019 reply Follow Share @Hemanth_13 why we cant have that reasoning for sigma star ? 0 votes 0 votes Hemanth_13 commented Jan 1, 2019 reply Follow Share Good question.. I think its because sigma is for input alphabet and our input alphabet will never contain $\epsilon$ 0 votes 0 votes shreyansh jain commented Jan 1, 2019 reply Follow Share $\sum^*$ is one of the languages from $L$, and it doesn't imply statement $2$ to be TRUE for every case. Rest is explained by @Hemanth_13 0 votes 0 votes HeadShot commented Jan 2, 2019 reply Follow Share @Hemanth_13 btw hemanth, when i looked closely your 1st comment, you said $L^+$ = { $\epsilon$ ,..} but $L^+$ it self means one or more occurrence then how that language will have epsilon at the 1st place, i am confused here. 0 votes 0 votes Please log in or register to add a comment.