# Is $E_{LBA}$ a Turing-recognizable language? [closed]

I know that $E_{LBA} = \{\langle M \rangle ~ \mid ~ L(M) = \emptyset \}$ is an undecidable language, but is it recognizable (recursively enumerable)? It seems that it's complement is recognizable since we could construct a Turing machine that enumerates all strings and checks if any belong to the language. If both $E_{LBA}$ and its complement were recognizable, then $E_{LBA}$ would be decidable, but it isn't, which leads me to think it isn't recognizable. Is this true?

## closed as unclear what you're asking by Yuval Filmus, David Richerby, Evil, hengxin, GillesDec 19 '16 at 10:57

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• That reasoning sounds good! – templatetypedef Dec 14 '16 at 14:54