$|a_n-\alpha|<2\epsilon$ valid as a definition of convergence? My question is concerning the proof below taken from Ivan Wilde’s notes on real analysis which are freely available on the web.
Is coming within $\large k\epsilon$ [$k\in\mathbb Z$] of a point as effective a proof of convergence as coming within $\large \epsilon$ of it?
For k$\epsilon$ can be made arbitrarily small.
Against - (and specifically regarding the question below) - since the reasoning always comes out at 2$\epsilon$ it cannot ever meet the definition of convergence

*

*but then you could just substitute and redefine the variable at the last step.

*is there a meaningful analogy between this and one of the better known proofs of the irrationality of $\surd 2$ whereby the numerator of $\frac{p}{q}$ is always even, and it is that which provides the contradiction.


THEOREM 
Suppose that $(a_n) [n ∈ N]$ is a sequence such that $a_n$ → α and also $a_n$ → β as n → ∞.
Then α = β, that is, a convergent sequence has a unique limit.
PROOF
Let ε > 0 be given.
Since we know that $a_n$ → α, then we are assured that eventually $(a_n )$ is within ε of α. Thus, $\exists\;N_1 ∈ N:  n > N_1\implies |a_n - α|<ε$
Similarly, we know that $a_n$ → β as n → ∞ and so eventually $(a_n)$ is within ε of β. Thus, $\exists\; N_2 ∈ N:  n > N_2\implies |a_n - α|<ε$
So far so good. What next? To get both of these happening simultaneously, we let $N = max { N_1 , N_2 }$.  Then n > N means that both $n > N_1$ and also $n > N_2$ . Hence we can say that if n > N then both
$| a_n − α | < ε$
and
$| a_n − β | < ε$.
Now what? We expand out these sets of inequalities. n > N (for example n = N + 1 would do). Then
Pick and ﬁx any
$α − ε <a_n < α + ε$
$β − ε < a_n < β + ε$
The left hand side of the ﬁrst pair together with the right hand side of the second pair of inequalities gives $α − ε < a_n < β + ε$ and so
$α − ε < β + ε$.
Similarly, the left hand side of the second pair together with the right hand side of the ﬁrst pair of inequalities gives β − ε < a n < α + ε and so
$β − ε < α + ε$.
Combining these we see that
$−2ε < α − β < 2ε$
which is to say that $\large| α − β | < 2ε$.
This happens for any given ε > 0 and so the non-negative number $| α − β |$ must actually be zero. But this means that α = β and the proof is complete.

It is totally correct actually you can fix $ε>0$;
then when applying the definition of convergence in coming steps set $ε'=ε/2>0$
then you will have at the end $ε/2+ε/2=ε$
actually  you can prove easily that for a given function of epsilon $f(ε)$ such that if epsilon goes to 0 , then $f(ε)$ goes to 0 .
if you have $(∃N∈N) n≥N⟹|a_n−α|<f(ε).$
then $(∀ε>0)(∃N'∈N) n≥N'⟹|a_n−α|<ε.$
