Is true that $L_1 = \{01^*0\}$ is $m$-complete in the class of decidable languages? Question: 


  Is true that $L_1 = \{01^*0\}$ is $m$-complete in the class of
  decidable languages?
  
  $L_1$ is defined as:
  
  $$L_1 = \{01^*0\} := \{01^n0: n \in \mathbb{N}\}$$


Definition of a $m$-complete language: 


  A language $L^*$ is $m$-complete in the class of decidable language,
  if for any decidable language $L$ we have that $L \leq_m L^*$ (that
  is, $L$ reduces to $L^*$).


I'm really new in this realm, so I'm quite lost about how to tackle this exercise.

I'm not sure if my approach is correct but I was trying to prove that the statement is not true by using a reduction from the Busy Beaver Function to $L_1$ (then, from this reduction and from $L_1$ being decidable we would have that BB problem is decidable, which is not possible).

Let $A$ be the algorithm deciding $L_1$ and $w \in L_1$ a word with $|w| = m$ ones. Now when running $A$ on $w \in L_1$ we keep track of the number of steps $t$ it made just before halting. Since $t = m|\Gamma|^m|Q|$, where $\Gamma$ is the set of tape symbols and $Q$ is the set of states. From here we can find the number $|Q|$ of states required for printing $|w| = m$ ones...

I stopped here because I feel there is already something wrong.

First, even is we find that number $|Q|$ for which $A$ performed $t$ steps before $A(w) = \text{Accept}$, that does not necessarily mean that $|w| = m$ ones could have been achieved with a TM with lesser number of states. Second, ''knowing'' the numbers of ones I'm determining the numbers of states and not the other way around, which is as $BB$ is defined.

Another alternative I was exploring was, on input $w$, to run all TM's $T_1, ... T_n$, that return $|w| = m$ ones. Then we record the input $w$ and the description of the machine $T_i,\ (i \in [1,n])$, that had the least number of states. If we do this $\forall w \in L_1$, we end up with a UTM $T'$, such that:

$$T' = \{ \langle T^*, w\rangle: |Q|_{T^*} = \mathrm{min}(|Q|_{T_1}, ..., |Q|_{T_n}\ \text{}\}$$

Then clearly $T'$ decides $L_1$ (just have to run it on input $w \in L_1$) and at the same time allows us to compute $BB(n)$ for a fixed $n \in \mathbb{N}$ (since $n$ is encoded in the description of $T^*$).

Probably here there are some flaws in my reasoning, but in this case, do I have to wait that, on input $w$, all those TM's reach a halt or can I assume that the first machine that halted is the one with the least number of states?

I'd appreciate any help.
Your approach can't succeed. You're asked whether every decidable language reduces to&nbsp;$L_1$ but your approach is to try to show that some undecidable language doesn't reduce to it. That would show nothing about decidable languages.

But are you sure that your definition is correct? Every language except $\emptyset$ and&nbsp;$\Sigma^*$ is $m$-complete by the definition you give.