A Turing Machine that exclusively accepts an infinite string

Question

While reading some of proofs in Computability Theory, I came up with following conclusion:

We can design a Turing Machine which exclusively accepts finite strings (obvious).

Now while trying the same for infinite string, I came up with the following conclusion :

There exist no Turing Machine which exclusively accepts infinite strings.

Proof : Since acceptance in context to Turing Machine is defined in terms of finite amount of time, therefore every time we say that an infinite string $w$ is accepted by a Turing Machine, we usually talk about searching a "finite pattern" in $w$. So while designing a machine for $w$ there exist a $w'$ which is finite and contains the same "finite pattern" which we try to find in $w$.

Example of "finite pattern" : A string that contains a 0. (Assuming Binary Strings)

So my question is that is my proof right? and if yes, is there a better way to prove this?

Answer 1

In contrast to the commenters, it is perfectly possible to define a Turing machine formalism whose input can be infinite. Let $\Sigma$ be an alphabet, and let $\#$ be a symbol not in $\Sigma$. We say that a Turing machine $T$ accepts a finite word $w \in \Sigma^*$ if it halts in an accepting state when the tape is initialized by $w \#^\omega$ (i.e., $w$ followed by infinitely many copies of $\#$; I assume for simplicity that the tape is only one-way infinite). Similarly, $T$ accepts an infinite word $w \in \Sigma^\omega$ if it halts in an accepting state when the tape is initialized by $w$.

We say that a language $L \subseteq \Sigma^* \cup \Sigma^\omega$ is $\omega$-computable (non-standard terminology) if there exists a Turing machine which halts on all inputs, and accepts exactly the strings in $L$. We say that $L$ is $\omega$-r.e. if there exists a Turing machine that accepts the strings in $L$, and never halts on strings not in $L$.

Under this definition, the language $\Sigma^*$ consisting of all finite words is not $\omega$-computable (as we show below), but is $\omega$-r.e. (exercise). It then follows that $\Sigma^\omega$ is not $\omega$-r.e., and in particular not $\omega$-computable (though we can show the latter directly).

Suppose that $T$ is a machine which halts on all inputs and accepts $\Sigma^*$. In particular, it halts on $\sigma^\omega$, where $\sigma \in \Sigma$ is arbitrary. Suppose that it halts after $N$ steps, at the rejecting state $s$. Then $T$ also halts on $\sigma^N$ (exercise), at the same state $s$, thus rejecting the finite word $\sigma^N$.

The same proof also shows directly that no Turing machine accepts $\sigma^\omega$.