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I am not sure of the validity of a proof I read about the convergence of sequences in certain space. Let $X$ be the set of all bounded or unbounded sequences of complex numbers with distance

$$d(x,y) = \sum_{j=1}^{\infty}\frac{1}{2^j}\frac{|\zeta_j - \eta_j|}{1+|\zeta_j - \eta_j|}.$$

Claim: We have that the sequence $(\zeta^{(n)}_j) = x_n \to x = (\zeta_j)$ iff $\zeta^{(n)}_j \to \zeta_j$ for all $j = 1, 2, \dots$.

To prove it, first suppose that $x_n \to x$ and fix $j$. We have $$\frac{|\zeta^{(n)}_j - \zeta_j|}{1+|\zeta^{(n)}_j - \zeta_j|} \le 2^j \cdot d(x_n,x) \to 0,$$ as $n \to \infty$. Therefore, for $\epsilon > 0$ sufficiently small it holds that for $n$ large enough that

$$|\zeta^{(n)}_j - \zeta_j| < (1+|\zeta^{(n)}_j - \zeta_j|)\epsilon,$$ which implies that $$|\zeta^{(n)}_j - \zeta_j| < \frac{\epsilon}{1-\epsilon} < 2\epsilon,$$ and hence $\zeta^{(n)}_j \to \zeta_j$ for all $j$.

But for convergence of a sequence, we require that for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $|\zeta^{(n)}_j - \zeta_j| < \epsilon$ when $n \ge N$, not just when "$\epsilon > 0$ is sufficiently small".

For example, if I take $\epsilon$ slightly smaller than $1$ then $$|\zeta^{(n)}_j - \zeta_j| < \frac{\epsilon}{1-\epsilon} < 2\epsilon,$$ certainly doesn't hold.

So how can this proof be valid when the expression doesn't hold for all $\epsilon >0$?

This proof is from page $2$, problem 3 of these notes.

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    Given $\varepsilon > 0$, pick an $N(\varepsilon)$ such that $\lvert \zeta_j^{(n)} - \zeta_j\rvert < \min \bigl\{\frac{1}{3}, \varepsilon\bigr\}$ for $n \geqslant N(\varepsilon)$. – 2017-02-07
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    A general principle is that in the definition of limit, one can restrict $\epsilon$ to any interval $(0,\epsilon_0)$ without changing the definition. That is, the conditions $$\forall\epsilon>0\quad\exists N\quad \forall n\geqslant N\quad d(x_n,x)\leqslant\epsilon$$ and $$\forall\epsilon\in(0,\epsilon_0)\quad\exists N\quad \forall n\geqslant N\quad d(x_n,x)\leqslant\epsilon$$ are **strictly equivalent**. – 2017-02-07
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    @DanielFischer Where are you getting $\text{min}\{\frac{1}{3},\epsilon\}$ from? How does this help with my counterexample which shows the proof doesn't hold if $\epsilon$ is slightly smaller than 1? – 2017-02-08
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    @Did That's a very nice property. Is there a simple way to show that this holds? – 2017-02-08
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    Sure, the implication from $\forall\epsilon$ to $\forall\epsilon<\epsilon_0$ is trivial. For the other implication, assume the second property holds and note $N_0$ some $N$ given by the second property for $\epsilon=\frac12\epsilon_0$, then use $N_0$ to show that the first property holds for every $\epsilon\geqslant\epsilon_0$ as well. – 2017-02-08
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    @Did Ok so if we assume the second property holds and there exists $N_0$ such that $d(x_n,x) < \epsilon \in (0, \epsilon_0)$ for all $n \ge N_0$ then for any $\gamma > \epsilon_0$ we have $d(x_n,x) < \epsilon < \gamma$ for $n \ge N_0$. Why did you use $\epsilon = \frac{1}{2} \epsilon_0$? – 2017-02-08
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    To guarantee to have some $\epsilon<\epsilon_0$. – 2017-02-08

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Note that $N$ that "works" for $\epsilon$ also works for $\epsilon'$ if $\epsilon'>\epsilon$. Therefore, it indeed suffices to consider only "sufficiently small" $\epsilon$.