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Is $L=\{\langle M\rangle\mid \text{$M$ is a Turing machine and $L(M)$ is uncountable}\}$ decidable?

My intuition is that it is not, but I'm not sure if Rice's Theorem applies in this case. If it is not decidable, how can I prove that using reducibility?

Discrete lizard
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1 Answers1

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This is somewhat of a trick question. What you are missing is that there are no uncountable languages over a finite (or even countable) alphabet. This should be enough information to answer it.

(I initially gave the entire answer away, but edited it out after considering a bit.)

Alexey Romanov
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