What is the sublanguage of L={ε, 1, 11, 111}

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TL;DR Summary: [Formal languages and automata]
Show all sublanguages of the language L={ε, 1, 11, 111} on Σ={0, 1}.

[Formal languages and automata]
Show all sublanguages of the language L={ε, 1, 11, 111} on Σ={0, 1}.

Is the answer {1}, {11}, {111}, {1, 11}, {11, 111}, {1, 11, 111}?

Or {ε}, {ε, 1}, {ε, 11}, {ε, 111}, {1, 11}, {1, 111}, {11, 111}, {ε, 1, 11}, {ε, 1, 111}, {ε, 11, 111}, {1, 11, 111}?
 
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If you wrote down the definition of sublanguage on ##\{0,1\}## more people could probably help.
 
Nico said:
Is the answer {1}, {11}, {111}, {1, 11}, {11, 111}, {1, 11, 111}?

Or {ε}, {ε, 1}, {ε, 11}, {ε, 111}, {1, 11}, {1, 111}, {11, 111}, {ε, 1, 11}, {ε, 1, 111}, {ε, 11, 111}, {1, 11, 111}?
In my limited understanding (from a long time ago), ε (the empty string) is just as valid as any other string.

So I’d say your first approach is incomplete and your second approach is correct.

However, I think you have some mistakes. The ones I spotted are:
You missed out {1, 111} from your first answer.
You missed out {ε,1, 11, 111} from your second answer.

There may be other omissions – I only checked quickly.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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