If $L$ is an infinite ($|L|=|\mathbb{N}| $) decidable language, prove that it contains:
a) An infinite subset that is not recognizable.
b) An infinite subset that is recognizable and not decidable.

For the (a) I considered all the subsets of $L$, i.e., $\mathbb{P}(L)$. Since $|\mathbb{P} (L)| = |\mathbb{R}|$, there exists $A \subset L$ that is not recognizable (for there are not enough Turing machines to recognize them all). If $A$ were finite, it would be decidable and hence recognizable, which would be a contradiction. Therefore $A$ is infinite and not recognizable.

For (b) I thought about $B=L\setminus A$. Then $B$ is recognizable because $A$ is not. But I'm not really sure.

  • 1
    $\begingroup$ If you’re not sure, try proving it. $\endgroup$ Commented Oct 28, 2020 at 6:13

1 Answer 1


Let $K$ be the set of indices of Turing machines which halt on the empty input. Consider the following language:

$$X = \{ 0 w : |w| \in K \} \cup \{ 1 w : |w| \notin K \}. $$

You can check that neither $X$ nor its complement are recognizable.

Therefore your proof idea doesn't work. Here is a different idea. Let $w_1,w_2,w_3,\ldots$ be the words in $L$, enumerated according length and then lexicographically (so if $L = \Sigma^*$, the order would be $\epsilon,0,1,00,01,10,11,\ldots$). The language

$$ Y = \{ w_i : i \in K \} $$

is recognizable but not decidable.

  • $\begingroup$ But I don't understand how are you defining Y $\endgroup$
    – user719961
    Commented Oct 28, 2020 at 23:17
  • $\begingroup$ Does it contain all the words whose indices correspond to a Turing machine which halts on the empty input? $\endgroup$
    – user719961
    Commented Oct 28, 2020 at 23:19

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