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.

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 $\varepsilon_0$.
