Just a very easy question that is making me wonder. Mind that, as a self-learner, even if this sound silly, I never really got an answer on this.
When we use $\varepsilon >0$ and we multiply or divide it, do we actually treat it (implicitly) as a positive integer or as an arbitrary real number such that $\varepsilon \in (0,1)$?
[I think that my confusion comes with the fact that $\varepsilon >0$ is usually presented as an arbitrary small real number that can be as close as you wish to $0$!]
In other words, does $\varepsilon/2$ makes the quantity smaller or bigger?
Any feedback is welcome. Thank you for your time.
PS: This is in some way related to a previous question of mine, where I use this at the end of the proof and I was really wondering about what I should really write down.
If all you say is that $\varepsilon > 0$, then you don't necessarily have that $\varepsilon$ is an integer (since most real numbers greater than $0$ are not integers), and you also do not know that $\varepsilon \in (0,1)$.
Regarding your second question: $2\varepsilon$ is bigger than $\varepsilon$ no matter what $\varepsilon$ is, as long as it's positive.
The reason one usually thinks about $\varepsilon$ as a small number is that in analysis, we often want to prove that "something is true for all $\varepsilon > 0$", and depending a bit on what that "something" is, it's usually the case that the smaller $\varepsilon$ is, the harder it is to prove what you want. Conversely, when you have proved this "something" and want to actually apply it, then the smaller $\varepsilon$ is, the more powerful a result you get out of it.
Here's a simple example: Say we want to show that $1/n \to 0$ for $n \to \infty$. That is, we want to show that for all $\varepsilon > 0$, there is a natural number $N$ so that $1/n < \varepsilon$ for all $n > N$.
Now say that we start by showing this for $\varepsilon = 60$. Can we find an $N$ so that $1/n < 60$ for all $n > N$? Yeah, that's completely trivial. Any natural $N$ will do the job.
On the other hand, if $\varepsilon$ had been really small, say $\varepsilon = 1/2^{60}$, then we actually need to think a little bit to cook up the necessary $N$. (Well, not that much, but hopefully you see the point.)