##### Subset of a language in Regular Expression

(1) Subset of a language in RE is always RE.

(2) A language is not RE is uncountably infinite.

statements are true/false and why?

(1) Subset of a language in RE is always RE.

(2) A language is not RE is uncountably infinite.

statements are true/false and why?

A regular language is not closed under subset operation.

and any finite language is regular. regular contains all finite languages and some infinite. so, a lang not RE is obviously infinite

^ how second one is true?

@sudsho

any finite language is regular and a FA can be designed for it. So a lang not regular is definately infinite.

language not regular is definitly infinite...fine

(2) A language is not RE is

uncountably infinite.there is no difference between countable infinite and uncountable infinite?(1) Subset of a language in RE is always RE.false..RE are not closed under subset operation

(2) A language is not RE is uncountably infinitefalse...we know that

sigma star is countable...alanguageis nothing but asubset of sigma star..every subset of a countable set is countable..henceall languages are countable...thoughset of languages which are not RE are uncountable....mind my word here.."set"