Why there is no Turing Machine that accepts the Diagonal Language?

Question

Given the diagonal language

$$L_d = \{ i : \sigma_i \notin L(M_i) \}$$

Where $M_i$ are all Turing Machines and $\sigma_i$ are all the words, if you put in in a Matrix like this:

$$\begin{array} {|c|c|c|c|c|c|c|} \hline & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 & \sigma_5 & ...\\ \hline M_1 & 1 & 0 & 1 & \dotsb & \dotsb & \dotsb \\ \hline M_2 & 0 & 0 & 1 & \dotsb & \dotsb & \dotsb \\ \hline M_3 & 1 & 0 & 1 & \dotsb & \dotsb & \dotsb \\ \hline M_4 & \vdots & \vdots & \vdots & 1 & \dotsb & \dotsb \\ \hline M_5 & \vdots & \vdots & \vdots & \vdots & \ddots & \dotsb \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \hline \end{array}$$

Then $L_d$ is represented by the numbers in the diagonal of the matrix. In class I was told that there is no TM that accept $L_d$, but I do not quite understand why is that, could somebody help?

PS: The above explanation was included because I did not know if this is called Diagonal Language in English, Spanish is my mother tongue.

Answer 1

In the following I assume that the matrix has an 1 in coordinates $(i,j)$ if $M_i$ accepts $\sigma_j$ and a $0$ otherwise (swapping all 1s and 0s does not affect the argument).

Consider the language $L = \{\sigma_i \in \Sigma^* : \sigma_i \not\in L(M_i) \}$ that contains all the words $\sigma_i$ such that the entry at coordinates $(i,i)$ of the matrix is $1$. I will first show that $L$ is not acceptable.

Suppose towards a contradiction that there is some Turing machine $M^*$ that accepts $L$. Since the set of Turing machines is enumerable, and the rows of the matrix are associated with an enumeration of all Turing machines, there must be some integer $i \ge 1$ such that $M_i = M^*$.

Then consider the word $\sigma_i$ and suppose that the entry at coordinates $(i,i)$ is $1$. From the way the matrix is constructed you know that $M_i$ accepts $\sigma_i$. However, by the definition of $L$, you also know that $\sigma_i \not\in L$. This implies that $M^* = M_i$ cannot accept $\sigma_i$. This is a contradiction.

If the entry at coordinates $(i,i)$ is $0$, then $M_i$ does not accept $\sigma_i$. However, $\sigma_i \in L$ implying that $M^* = M_i$ accepts $\sigma_i$. Again, this is a contradiction.

Going back to your language $L_d$: if $L_d$ were acceptable then so would be $L$. Suppose towards a contradiction that there is a Turing Machine $M$ that accepts $L_d$. To accept $L$ proceed as follows:

• find the index $i$ such that $\sigma_i = x$. This can be done by enumerating the words in $\Sigma^*$.
• Simulate $M$ with input $i$.
• If $M$ with input $i$ halts and accepts, accept.

Since this is a contradiction, $L_d$ cannot be acceptable.