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

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.

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

The reason is that the typical definition of a Turing machine is that its representation should contain a non-empty set of states and a non-empty tape alphabet.

In order to represent those, you typically would need a non-empty string.

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, $\varepsilon \notin E_{TM}$.

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