Convergence Proof: Difference within Function of Epsilon Here's the problem, and then my proposed proof:

Suppose that $\phi : (0, \infty) \mapsto (0, \infty)$ is a function such that $\lim_{\epsilon \to 0}\phi(\epsilon) = 0.$ Let $a_1,a_2,\dots$ be a sequence in $\mathbb{R}^n$. Show that this sequence converges to $a$ if and only if for any $\epsilon > 0$, there exists $N$ such that for $n > N$, we have $|a_n-a|\leq \phi(\epsilon).$

(from Vector Calculus, Linear Algebra, and Differential Forms (Hubbard and Hubbard))
My proof is for the first statement (the if, not the only if).

First, to avoid confusion, the argument of the function $\phi$ will be renamed to $\epsilon'$.By the definition of the limit, the statement that $\lim_{\epsilon \to 0}\phi(\epsilon) = 0$ is the same as saying that $$\forall\epsilon >0, \exists \delta > 0 \text{ such that } |\epsilon'| \leq \delta \implies |\phi(\epsilon')| \leq \epsilon $$
By assumption, $|a_n - a| \leq \phi(\epsilon')$, and from the statement above, we get
$$|a_n - a| \leq \phi(\epsilon') \leq \epsilon$$
We want $\phi(\epsilon') \leq \epsilon'$. However, since it must be true for all $\epsilon$ and all $\epsilon'$ greater than 0, the two can be taken to be equal. Therefore,
$$|a_n-a| \leq \epsilon'$$

I'm unsure about the final step, where I say that the unprimed and primed epsilon can be taken to be equal. Is this correct?
 A: The if case
Assumptions


*

*$\lim_{\epsilon' \to 0} \phi(\epsilon') = 0$, i.e. for any $\epsilon>0$ there exists $\epsilon''>0$ such that for $0<\epsilon'<\epsilon''$ we have $0 \leq \phi(\epsilon')<\epsilon$

*for any $\epsilon'>0$ there exists $N$ such that for $n>N$ we have $0 \leq |a_n−a| \leq \phi(\epsilon')$


To show
$\lim_{n \to \infty} a_n = a$, i.e. for any $\epsilon>0$ there exists $N$ such that for $n>N$ we have $|a_n−a| < \epsilon$
Proof
Take $\epsilon>0$.
By assumption 1 there exists $\epsilon''>0$ such that for all $0 < \epsilon' < \epsilon''$ we have $0 \leq \phi(\epsilon') < \epsilon$. For such $\epsilon'',$ let $\epsilon'=\epsilon''/2$. Then $0 \leq \phi(\epsilon') < \epsilon.$
By assumption 2 there now exists $N$ such that for $n > N$ we have $0 \leq |a_n-a| \leq \phi(\epsilon')$. But by the previous paragraph we have $\phi(\epsilon') < \epsilon$ so $0 \leq |a_n-a| < \epsilon$.
Thus, for any $\epsilon>0$ there exists $N$ such that for $n > N$ we have $|a_n-a| < \epsilon$.
