Auxiliary Epsilon - is it mathematically rigorous if it's larger than epsilon? Assume $a_n \rightarrow L$. Thus, applying the standard definition, $\forall \epsilon > 0, \exists N\in\mathbb{N},\, \forall n\geq N, |a_n-L| < \epsilon$.
I'm in an introductory real analysis class, and my professor will sometimes use an "auxiliary epsilon" $\epsilon_0$ and show that $|a_n - L| < \epsilon_0$ where $\epsilon_o$ is defined in terms of $\epsilon$.
I was having a debate with my friend about the relationship between $\epsilon_0$ and $\epsilon$. 
I believe that for this proof to be rigorous, $\epsilon_0$ must be less than or equal to $\epsilon$ for it to be sufficient to show  $|a_n - L| < \epsilon_0$ instead of $|a_n - L| < \epsilon$. So for example, arriving at $|a_n-L| <  \epsilon_0 = c*\epsilon$ for $c > 1$ would not be mathematically rigorous.  Like if you chose epsilon to be $.1$, I don't see how it's enough to say that $|a_n-L|$ must be less than $.2$, if $c$ were $2$ here.
However, my friend argued that because it’s an arbitrary positive $\epsilon$, as long  as the statement is universal, it doesn’t matter because you can take a smaller epsilon to fit your range. So he's saying that in the case where $\epsilon = .1$ and $\epsilon_0 = .2$, you can choose $\epsilon =.05$ so $\epsilon_0 = .1$ which satisfies the original $\epsilon = .1$.
Which is correct? I presume I'm looking at it like $\epsilon-\delta$ proofs of limits, where you must specify a $\delta$ for which all corresponding function values are within $\epsilon$ of the limit. If $\delta$ is too wide and there's some value $x$ for which $|x-a|<\delta$ and $|f(x)-L|>\epsilon$, then the proof is invalid. I'm thinking the same logic would apply here.
 A: Your friend is correct in this case. Keep in mind that in an $\varepsilon-\delta$ proof, we are never really choosing an explicit value for $\varepsilon$; rather, we are saying that for any arbitrarily small $\varepsilon$, our sequence of $\{a_n\}$ is within $\varepsilon$ of $L$ eventually. If $\varepsilon_0=c\varepsilon$, then if $\varepsilon$ is taken to be arbitrarily small, then so is $\varepsilon_0$.
A: Ben's answer addresses the heart of the matter, but I want to complement it with an example.
Say we proved that for any $\varepsilon>0$ there is $N_\varepsilon\in\mathbb{N}$ such that $n\ge N_\varepsilon$ implies $|a_n-L|<c\varepsilon$, for some fixed constant $c$.
Now we can use this result to prove what we really want.
Indeed, fix $\varepsilon>0$ and choose $N:=N_{\varepsilon/c}$.
Then $N$ satisfies
$$n\ge N \implies |a_n-L|<c\left(\varepsilon/c\right)=\varepsilon.$$
Hence we have shown precisely what is required by the definition of sequence convergence. This same "trick" works for other limits, continuity, etc.., which is why authors will usually just end their proof without its inclusion (or mention).
Here's an explicit example:

If $a_n\to a$ and $b_n\to b$, then $a_n+b_n\to a+b$.

Proof:
Fix $\varepsilon>0$. Since $a_n\to a$, there exists $N_1\in\mathbb{N}$ such that $|a_n-a|<\varepsilon$ whenever $n\ge N_1$. Similarly, from $b_n\to b$ we can find $N_2\in\mathbb{N}$ such that $|b_n-b|<\varepsilon$ whenever $n\ge N_2$.
Thus, taking $N:=\max\{N_1,N_2\}$, we obtain
$$
n\ge N \implies |a_n+b_n-(a+b)|\le|a_n-a|+|b_n-b|<2\varepsilon. \tag*{$\square$}
$$
In the proof above we applied the definition of $a_n\to a$ and $b_n\to b$ for $\varepsilon$ to obtain an $N$ that "works" for $2\varepsilon$. Alternatively, we could have applied the definition of $a_n\to a$ and $b_n\to b$ for $\varepsilon/2$ to obtain an $N$ that "works" for $\varepsilon$. Let's see how this would look.
Proof:
Fix $\varepsilon>0$. Since $a_n\to a$, there exists $N_1\in\mathbb{N}$ such that $|a_n-a|<\varepsilon/2$ whenever $n\ge N_1$. Similarly, from $b_n\to b$ we can find $N_2\in\mathbb{N}$ such that $|b_n-b|<\varepsilon/2$ whenever $n\ge N_2$.
Thus, taking $N:=\max\{N_1,N_2\}$, we obtain
$$
n\ge N \implies |a_n+b_n-(a+b)|\le|a_n-a|+|b_n-b|<\varepsilon. \tag*{$\square$}
$$
