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c is answer. till recursive enumerable we have enumerator

A language is Turing recognizable if and only if some enumerator enumerates it.

but for non re no enumerator

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@Anu007 sir,

Turing machines are countable

hence RE also countable and hence we can enumarate REL as well .

and hence I think ans should be D

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i also thought D as answer .....set of all recursive enumerable language=set of all turing machine which are countable ..but given answer as C . but set of all language over 0,1 is also having enumeration method .

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Ohh sorry i read alphabets!

yes option C is true!

set of all languages 2^sigma* is uncountable!

we can prove it by diagonalization method.