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Use the following lemma and diagonalisation:

Lemma: Let $M$ be a linear bounded automaton. Then, there is a lba $M'$, that accepts the same language and halts on every input.

($\mathsf{DSPACE}(n)$ is all languages accepted by some linear bounded automaton.)

I think, the fact that all languages form $\mathsf{DSPACE}(n)$ are decidable is given by the lemma, isn't it? But how do I prove existence of some decidable language not in $\mathsf{DSPACE}(n)$? I guess I'd need some language that needs bigger and bigger tapes to be recognized so that no lba can do it...is this the right idea?..

Thanks

Ettore
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  • E is the decidable languages... – Ettore Oct 26 '20 at 08:05
  • Depending on your source $\subset$ can mean different things. Commonly it means $\subseteq$, which is subset or equal. In this case you are done with your observation that every $\mathsf{DSPACE}(n)$ language is decidable. Finding a decidable language not in $\mathsf{DSPACE}(n)$ is only necessary, if $\subset$ is meant as $\subsetneq$, ie. denotes proper subsets. – Jonas Linssen Oct 26 '20 at 08:23
  • thank you, yes I should have been more precise, we need to prove proper subset – Ettore Oct 26 '20 at 11:15

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