This question is about the Turing computability of the $\epsilon-N$ definition of a limit of an infinite sequence $S$. First, a proposition:

There cannot exist a Turing Machine $M$ which, given a program $P_S$ whose output is the sequence $S$, has output $$ M(P_S) = \left\{ \begin{array}{rl} \operatorname{true} & \text{if}\,\lim(S)\, \text{exists} \\ \operatorname{false} & \text{otherwise.} \end{array}\right. $$

In other words, a Turing Machine cannot decide if $\lim(S)$ exists. A short argument as to why this is the case is below.

My question is: what are the consequences of having such a fundamental definition being undecidable? Is there another definition which is decidable?

It seems to me that most mathematical definitions are decidable. My feeling is that definitions should be decidable.

Argument: A Turing Machine that decides the existence of $\lim(S)$ could be used to solve the Halting Problem in the following way.

Denote $P_S\oplus0$ as the program that runs $P_S$ and upon observing the termination symbol appends an infinite number of zeros. Similarly $P_S\oplus 01$ is the program that appends an infinite alternating sequence $01010101\ldots$.

Then $P_S$ halts iff ($M(P_S\oplus 0)$ and not $M(P_S\oplus 01)$).

Hence the existence of limits, for sequences given by algorithms, cannot be decided by a Turing Machine.

This question is about the Turing computability of the $\epsilon-N$ definition of a limit of an infinite sequence $S$. First, a proposition:

There cannot exist a Turing Machine $M$ which, given a program $P_S$ whose output is the sequence $S$, has output $$ M(P_S) = \left\{ \begin{array}{rl} \operatorname{true} & \text{if}\,\lim(S)\, \text{exists} \\ \operatorname{false} & \text{otherwise.} \end{array}\right. $$

In other words, a Turing Machine cannot decide if $\lim(S)$ exists. A short argument as to why this is the case is below.

My questions are:

1. is there another (perhaps weaker) definition of limit which is decidable?

2. a definition is "a statement of the exact meaning". Does an undecidable concept have an exact meaning? Perhaps undecidable statements are less suited to be definitions than decidable statements.

Argument: A Turing Machine that decides the existence of $\lim(S)$ could be used to solve the Halting Problem in the following way.

Denote $P_S\oplus0$ as the program that runs $P_S$ and upon observing the termination symbol appends an infinite number of zeros. Similarly $P_S\oplus 01$ is the program that appends an infinite alternating sequence $01010101\ldots$.

Then $P_S$ halts iff ($M(P_S\oplus 0)$ and not $M(P_S\oplus 01)$).

Hence the existence of limits, for sequences given by algorithms, cannot be decided by a Turing Machine.

This question is about the Turing computability of the $\epsilon-N$ definition of a limit of an infinite sequence $S$. I hope this generates some interesting discussion. My proposition is:

There cannot exist a Turing Machine $M$ which, given a program $P_S$ whose output is the sequence $S$, has output $$ M(P_S) = \left\{ \begin{array}{rl} \operatorname{true} & \text{if}\,\lim(S)\, \text{exists} \\ \operatorname{false} & \text{otherwise.} \end{array}\right. $$

In other words, a Turing Machine cannot decide if $\lim(S)$ exists.

My argument is that such a Machine could be used to solve the Halting Problem in the following way.

Denote $P_S\oplus0$ as the program that runs $P_S$ and upon observing the termination symbol appends an infinite number of zeros. Similarly $P_S\oplus 01$ is the program that appends an infinite alternating sequence $01010101\ldots$.

Then $P_S$ halts iff ($M(P_S\oplus 0)$ and not $M(P_S\oplus 01)$).

Hence the existence of limits, for sequences given by algorithms, cannot be decided by a Turing Machine.

This question is about the Turing computability of the $\epsilon-N$ definition of a limit of an infinite sequence $S$. First, a proposition:

There cannot exist a Turing Machine $M$ which, given a program $P_S$ whose output is the sequence $S$, has output $$ M(P_S) = \left\{ \begin{array}{rl} \operatorname{true} & \text{if}\,\lim(S)\, \text{exists} \\ \operatorname{false} & \text{otherwise.} \end{array}\right. $$

In other words, a Turing Machine cannot decide if $\lim(S)$ exists. A short argument as to why this is the case is below.

My question is: what are the consequences of having such a fundamental definition being undecidable? Is there another definition which is decidable?

It seems to me that most mathematical definitions are decidable. My feeling is that definitions should be decidable.

Argument: A Turing Machine that decides the existence of $\lim(S)$ could be used to solve the Halting Problem in the following way.

Denote $P_S\oplus0$ as the program that runs $P_S$ and upon observing the termination symbol appends an infinite number of zeros. Similarly $P_S\oplus 01$ is the program that appends an infinite alternating sequence $01010101\ldots$.

Then $P_S$ halts iff ($M(P_S\oplus 0)$ and not $M(P_S\oplus 01)$).

Hence the existence of limits, for sequences given by algorithms, cannot be decided by a Turing Machine.