I am about 99.999% sure this construction also works with topologies and the widgets that make topologies, but I'm too tired to write out the details. There may be some category-theoretic explanation of which properties play nice with this construction, but I don't actually know any category theory.

Elements of S are in $C^\infty$. Its something I know but I dont know why I got it wrong... thanks!

Eric: Thank You very much for the detailed explanation. However, I am not convinced about the additive and multiplicative inverses to make S a field, given my definition of S.

But wait... suppose you come up with a suitable explanation to my comment above, there is something in the next part that I do not follow. What was that thing about homomorphism/isomorphism? Does not isomorphism preserve the structure imposed on the set and bring about the same effect on both the sets? What do you mean by your statement regarding $\phi o \sigma$ ?

@Manasi: Existence of inverses is straightforward. The inverse of $s$ is $\sigma^{-1}\big(\phi^{-1}\big(\phi(\sigma(s))^{-1}\big)\big)$, or replace $s^{-1}$ with $-s$ for additive; the inverse is well defined because $\phi(\sigma(s))$ is a rational number. This construction also allows for all three generalizations. // Yes, $S$ as I have constructed it is isomorphic to the rational numbers. It has the same structure of addition and multiplication. $\phi\circ\sigma$ is simply the bijection which exhibits the desired isomorphism; it is worthwhile to check that it is indeed a homomorphism

This does not make sense. Say $\sigma(s)= 5$, say. Let $\phi(5) = k$, say where k is some rational number. Now since $k\neq 0$ , you have $\frac{1}{k}$ is again a rational number which may not be a positive integer and then one is in trouble since $\sigma^{-1}(\frac{1}{k})$ would not exist!

@Manasi: I think you saw an older version of the comment; it now correctly says $\sigma^{-1}\phi^{-1}$ instead of $\phi^{-1}\sigma^{-1}$. Now the concern is removed because $\phi^{-1}(1/k)\in\Bbb Z^+$ and so you can take $\sigma^{-1}$ of it.

Okay, thank You. I will take sometime to go through the whole thing again.

There is another poster who has imposed an operation through lcm to make S into a semi-group.

Yes, Gerry's construction is identical to mine, with $\phi$ the identity and $\sigma(s_n)=n$.

Of course, since the image of $\phi\circ\sigma$ is a semigroup rather than a field, we can only get the semigroup axioms from the morphism.

@EricStucky: Hmm... I just need sometime to mull over what you have written...

No problem :)

yeah, thats right reg semigroup

I will get back tomorrow...and post here if I struggle but if I get it, then I'll be fine.

sounds good

its late in the night now....thank you and bye

Just start your message with @EricStucky so that I get a notification.

good night :)

yeah I will do that and post here.

good day to you :)