According to wikipedia, every finite set is computable. Definition: set $S \subset N$ is computable if there exists an algorithm which defines in finite time if a given number $n$ is in Set.Question: what is wrong with this counter-example:given some $TM$$S \subset N$Lets assume $S$ could contain only $0$, i.e., either $S = \{0\}$ or $S = \emptyset$if a given $TM$ halts then $S=\{0\}$ otherwise $S=\emptyset$So set $S$ is finite, but not computable, since we cannot compute if a given $TM$ halts.What is wrong above?
Why every finite set is computable?
computability;turing machines
A similar question was answered here:How can it be decidable whether $\pi$ has some sequence of digits?We can ignore the question if a given TM halts as it is irrelevant what the question actually is, let's just name the condition C.If C is true, then the correct algorithm is if n == 0 return true else return false. If C is false, the correct algorithm is return false. Whether C is true or false, one of these two algorithms is correct for every n, so such an algorithm exists.Additionally, does a given TM halt is computable for the same reason - the correct algorithm is either return true or return false. What is not computable is a function that answers does this TM halt for any TM.