# Does $E_{TM}$ accpets the empty word $\varepsilon$?

Let $$L = E_{TM} = \left\{ \left | M \text{ is a TM and L(M)} = \emptyset \right\}$$.

Does $$L$$ accepts the empty word $$\varepsilon$$?

In other words, is $$\varepsilon \in L$$

I'm a little bit confused by this.

My intuition says it doesn't since the empty set rejects every input.

• The answer depends on how you encode Turing machines. Which Turing machine does the empty string represent, if any? – Yuval Filmus Apr 7 '19 at 6:35
• I think it would be reasonable to assume that no Turing machine has the empty string as a representation. – Pål GD Apr 7 '19 at 7:29
• @PålGD Sometimes we would prefer that any string represent some Turing machine. – Yuval Filmus Apr 7 '19 at 9:07

I would answer that no, $$\varepsilon \notin L$$.
However, as Yuval points out, you are free to define your own Turing Machine description language, and if you were to say "We use $$\varepsilon$$ to denote the "empty" TM (with alphabet 0), the TM that runs forever", you are free to do so, in which the answer becomes Yes.
I would go with No, $$\varepsilon \notin E_{TM}$$.