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Why all the properties of a sequence or a series or a sequence of functions or a series of functions remain unchanged irrespective of which of $\mathbb N$ & $\omega$ we are using as an index set? Is it because $\mathbb N$ is equivalent to $\omega$?

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Actuallly $\Bbb N$ is equal to $\Bbb N\cup\{0\}$, in my book. It may be worth while to state explicitly that you are assuming the convention that $0\notin\Bbb N$ as you obviously are. –  Marc van Leeuwen Dec 16 '12 at 10:31
    
@MarcvanLeeuwen That's always annoyed me. Every teacher I've ever had defined $0\notin\mathbb N$ and so I'm used to that, but we already have perfectly good notation for both sets: $\mathbb N$ for natural numbers and $\omega$ for whole numbers. I'm borrowing $\omega$ from set theory here (as the first limit ordinal), but it is conveniently equal to $\mathbb N\cup\{0\}$, and it looks like a W to boot. Also available alternative notation is $\mathbb Z^{\geq 0}=\omega$ and $\mathbb Z^+=\mathbb N$. –  Mario Carneiro Dec 16 '12 at 10:48
    
@MarioCarneiro: The notation for the set of whole numbers is $\mathbb Z$. The set of natural numbers is $\omega=\mathbb N=\{0,1,2,3,\ldots\}$. –  Henning Makholm Dec 16 '12 at 11:48
    
Also related math.stackexchange.com/questions/150575/… –  Asaf Karagila Dec 16 '12 at 11:54
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@BrianM.Scott I would gladly switch my notation if I knew that everyone was on the other side, but this is just one of those bad notation peculiarities with no consensus. See en.wikipedia.org/wiki/Natural_number for an overview of the history and arguments for each side and a few unambiguous alternative notations. One clear one which I use sometimes is $\mathbb N-\{0\}$ vs. $\mathbb N\cup\{0\}$, since it doesn't need any new symbols and works with both systems. –  Mario Carneiro Dec 16 '12 at 20:59

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up vote 6 down vote accepted

There is a trivial 1-1 correspondence from $\mathbb N=\{1,2,\dots\}$ to $\omega=\{0,1,2,\dots\}$ given by $f(n)=n-1$. This mapping is a bijection, an order isomorphism, an isometry, and preserves limits, in the sense that $f(n)\to\infty$ if $n\to\infty$ (although it would be hard not to satisfy this). Thus practically every property we care about is preserved when we switch from one index set to the other, almost to the point that we don't need to pay attention to which one we are working with.

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+1 for "would be hard" –  Hagen von Eitzen Dec 16 '12 at 11:36
    
@HagenvonEitzen Of course. Hard as in "I can prove it's impossible". :) –  Mario Carneiro Dec 16 '12 at 11:43

It is because $\omega$ and $\mathbb N$ are just different names for the same set. Their members are the same, and so by the Axiom of Extensionality they are the same set.

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