Is the infinite language unrecognizable in a Turing machine?

Question

This question is building up on an older one, here.

But now let's say we keep $Σ=\{0,1\}$. Is the TM that accept anys ($1^x \mid x \gt 0$) recognizable?
That means 1, 11, 11111, 1111111, and so on are all accepted.

I believe the TM is unrecognizable. This is because for it to be recognizable, we would have to halt. Yet, with no upper limit, we can have $1^\infty$ - meaning we will never halt. Does this work?

The above was an example of a language, didn't think it would be recognizable. A better way to phrase my question is: Is there a language of any number of $1$s over $\Sigma=\{0, 1\}$ that you can make unrecognizable?

Answer 1

I'm a bit confused by your question: you're asking if the Turing machine is recognizable, but I think you mean to ask if the language $\{1^x \mid x \in \mathbb{N}\}$ is recognizable.

A language is recognizable if and only if we can build a Turing machine that accepts every string in the language, and does not accept any string not in the language. And we can indeed build a Turing machine that does this!

Algorithm:
    Check the number under the head.
    If it's 0, fail.
    If it's the end of the string, accept.
    If it's 1, move to the right and repeat.

The key is, while there's no upper limit, $x$ has to be a natural number—and $\infty$ is not a natural number. In other words, while $x$ can be arbitrarily large, it has to have some finite value. It can't actually be $\infty$.

Answer 2

Languages are sets of finite strings. Every input to a Turing machine is a finite string. $1^\infty$ is a thing, but not in this model of computation (and usually we're more specific about what infinity we're talking about).

Answer 3

A few answers has addressed the confusion about the length of a word being infinite.

Here I would like to address the following question.

A better way to phrase my question is: Is there a language of any number of 1s over $\Sigma=\{0,1\}$ that you can make unrecognizable?

Note that if a language of any number of 1s is unrecognizable, then it is unrecognizable over any alphabet that contains 1.

That question has been answered positively here. I would like to add another famous "constructive" example, busy beaver sequence, a.k.a. Rado's sigma function. Let $\sigma:\Bbb N\to\Bbb N$ be that function. We know that $$\sigma(0)=0,\ \sigma(1)=1,\ \sigma(2)=4,\ \sigma(3)=6,\ \sigma(4)=13, \\\sigma(5)\ge4098,\ \sigma(6)\ge 1.29*10^{865},\cdots. $$

We have not determine $\sigma(5)$ yet!

The language for $\sigma$ is $$\{1^{\sigma(n)}: n\in \Bbb N\}=\{\epsilon, 1, 1111, 111111, 1111111111111, \cdots\},$$ which is undecidable.

Answer 4

$1^\infty$ is not an actual word in the language, since $\infty$ is not a natural number - even though the language contains infinitely many words, each word has a finite, arbitrarily large length $n$.

A language is decidable if there is a Turing Machine can decide it in a finite number of steps for each input. That's obviously true here - there is no word that will take infinitely long to process.

The above was an example of a language, didn't think it would be recognizable. A better way to phrase my question is: Is there a language of any number of 1s over $Σ=\{0,1\}$ that you can make unrecognizable?

You mean an undecidable unary (consisting only of one symbol) language? Sure there is!

Take any language $L \in \{0,1\}^*$ that you already know to be undecidable. (For example, a language that encodes all halting Turing Machines.) Modify it a bit by adding "1" to the start of each word, to keep leading zeros.

Then let your language $L'$ consist of the word $1^n$ for every $n$ whose binary representation is in $L$. ~~As each word in $L$ can be easily turned into a word in $L'$~~*, a machine that decides $L'$ could be used to decide $L$. As $L$ is undecidable, so is $L'$.

Edit: *To be more precise: "As each binary word $w$ beginning with 1 can be converted into a word consisting of 1s $w'$ such that $w\in L$ iff $w' \in L'$..."