Turing Machines Deciadability

Question

L ={ T | e(T) belongs to L(T) } , e(T) is the code for turing machine?

The answer given is RE but not REC ,but cant we apply Rice theorem as follow-

Tyes={Any machine accepting its code} Tno={Machine accepting code of Tyes + a }  //Tno is not accepting its own code

Tyes will become subset of Tno.?

Anyhow we can prove it as RE by simply checking that when e(T) will be applied to T,then members will say say and for Non members it may or may not halt.But where am i wrong in Rice's theorem proof?

Comments

it is REL but not REC,

here property of language L is that "e(t) ϵ L(T)", now if i say some turing machine, e(t)= 11010 then there exist one language T_yes={11010, 00101, 10100}, since 11010 is present in T_yes, and another language Tno={ }, since 11010 is not present in T_no...

since there exist T_yes and T_no, we can say it is not REC, but you can see clearly that T_yes is not proper subset of T_no is this example or can not be in any other example...threfore it qualify to be REL

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Why cant 11010 can belong to the language of machine Tno. I i have a Tno as {11010 ,0},then why is it invalid for Tno?

Lets us say Tno code is :-1111, now Tno is not having its own code in its language but it is having code of Tyes.Please explain this