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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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    @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 http://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.2012-12-16

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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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    @HagenvonEitzen Of course. Hard as in "I can prove it's impossible". :)2012-12-16
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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.