True or False: If $A \subseteq \{0,1\}^* \Rightarrow A^*$ is semi-decidable
It is not true that every language is recursively enumerable ("type 0").

Proof 1. Every RE language is semi-decided by some Turing machine. There are only countably many Turing machines (each one can be coded as a finite string, or an integer) but there are uncountably many languages.  Therefore, there are uncountably many languages that are not RE.

Proof 2. If a language and its complement are both semi-decidable, they are, in fact, both decidable.  The halting problem is semi-decidable. If the complement of the halting problem were semi-decidable then both it and the halting problem would be decidable. But we know that the halting problem is not decidable. Therefore, the complement of the halting problem is not semi-decidable.