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If I have the following set $E_n$ where n denotes a natural number. What does putting it in subscript mean ?

Thanks

Edit :

Difference of two natural integers:

For all integers $m,n \in \mathbb N$ which satifies $m>n$, we have $m-n \in\mathbb N$. Indeed, for all $n \in \mathbb N$, the set $$E_n=\{m-n \in \mathbb N : m \in \mathbb N, m>n\}$$ contains ($n+1)-n = 1+ (n-n) = 1+ 0 = 1$ with $n+1>n+0 = n$. Furthermore, if $k$ is element of $E_n$, there exists $m \in \mathbb N$ with $m>n$ such that $$k+1= (m-n)+1= (m+1)-n$$ with $m+1 \in \mathbb N$ satisfying $m+1>n+1>n+0=n$ where $k+1 \in E_n$

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    That it's dependent on $n$? It really depends on what you're told that $E_n$ is. For example, if $E_n = \{1, \dots, n\}$, then $E_1 = \{1\}$, $E_2 = \{1, 2\}$, etc. – 2017-01-14
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    Sorry, here's the full set : E_n = ( n element of naturals :for all m elements of naturals, m*n is also element of naturals) I guess it's the second part (formally), right ? It's part of some proposition that I have to prove. – 2017-01-14
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    Well, $E_n=\Bbb N$, for all $n \in \Bbb N$ – 2017-01-14

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Intuitively, it means that $E_0$, $E_1$, $E_2$ and so forth are names of different sets -- though presumably ones that you're going to say something general about later on.

Formally, it is just a different notation for saying you have some function $E:\mathbb N\to\text{something}$, where you're declaring that by convention the function values written $E_n$ rather than $E(n)$.

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    @copper: It not an either/or -- I'm describing two ways to think about the _same_ common notation. – 2017-01-14
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    So how does this fit exactly with my set ? I would imagine that I have several sets where my n takes on different values of naturals that are being multiplied by m ? – 2017-01-14
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    @copper: The notation $E_n$ implies you're supposed to have several different sets, but the definition you're quoting does not really manage to define that -- its mention of $n$ is _bound_ by the set builder notation. There must be a typo or misquote somewhere; I think you have to show us some broader context before we can hope to unravel what your exercise is trying to say. – 2017-01-14
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    Ok, I added the full proposition in the original question! – 2017-01-14
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    @copper: You forgot "$m-$" at the beginning of the set builder when first quoting the definition. Now it seems to make some sense, but the _result_ is still that $E_0, E_1, E_2$ and so forth all end up being the _same_ set $\{1,2,3,\ldots\}$. Presumably the text has some point with doing that, though what it is I cannot guess. – 2017-01-14
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    So, just to reformulate, $E_0$, etc. are all the values taken by n for each of the different sets, right ? And yeah, I'm not sure myself why it's written this way. I don't see what new information it brings if it didn't appear this way. – 2017-01-14
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    @copper: No, the meaning would be something like $$ E_0 = \{m\mid m\in\mathbb N, m>0 \} \\ E_1 = \{m-1\mid m\in\mathbb N, m>1\} \\ E_2 = \{m-2 \mid m\in\mathbb N, m>2 \}$$ and so forth. – 2017-01-14