TOC(Undecidability)

Question

L1 = { <M> | M halts on $\epsilon$ }

L2 = { <M> | $\epsilon$   $\in$ L(M) }

Which one is RE or not RE

Comments

got it thanks sir:)

same algorithm we can use for L1 also ..just given epsilon as input ? so both are re but not rec

sir in first one , L1 it is given TM halts on ϵ ...that doesn't mean ϵ should belong to L(TM) right ? TM may halt in non final state ..in L2 we can atleast apply 1st property of rice theorem but in L1 we can't apply 1st also

M I right sir?

---

Both the languages mentioned in the question, L1 and L2 are RE. only the difference is, in the case of L1, the TM has to halt either in final state or non--final state, while in L2, TM will halt in a final state, means, TM will accept epsilon as a member of the language.

if we slightly change the language L2, as L2 contains ONLY epsilon as the member of the language L2, then it is NOT RE.

---

not understood ,Please anyone Explain  clearly?