Really simple question about a specific Cauchy sequence in a generic normed space. Context. Let $X = (X,\|\cdot\|)$ be a normed space and $R:(X,\|\cdot\|) \longrightarrow (\Bbb K^n , \|\cdot\|_\infty)$ a linear and topological isomorphism between $X$ and $\Bbb K^n$ (here, $\Bbb K \in \{\Bbb R, \Bbb C\}).$
Question itself. Take any Cauchy sequence $(x_k)_{k \in \Bbb N} \subset X.$ Thus, by definition of Cauchy sequence,
$$ \forall \epsilon >0, \exists N = N(\epsilon) \in \Bbb N\, : \, \forall n,m \in \Bbb N, \, n,m > N \Rightarrow \| x_n - x_m \| < \epsilon. $$
Now, one easily notes that
$$ \| R(x_n) - R(x_m)\|_\infty = \| R(x_n-x_m)\|_\infty \leqslant \| R \| \, \|x_n-x_m\| < \|R\| \, \epsilon. \quad (n,m > N) $$
Note that I used linearity of $R$, aswell as the fact that $R$ is continuous and thus bounded. And here urges my question:
I certainly can't say that $\|R\|\epsilon < \epsilon$ because this inequality is valid only when $\|R\| < 1.$ Normally, what people do is to define initially $\|x_n-x_m\| < \frac{\epsilon}{\|R\|}$ but again, this doesn't makes sense in my head because $\epsilon < \frac{\epsilon}{\|R\|}$ isn't valid for all values of $\|R\|$ (I am pretty sure an operator can have a norm between $0$ and $1$).
At the same time, my intuition is that the sequence $(R(x_k))_{k \in \Bbb N}$ is also Cauchy (in $\Bbb K^n$) but I am having some troubles with this $\epsilon$ thing... Any help would be apreciatted.
 A: (You should rename the dimension of your spaces to something that's not $n$, $m$ or something already taken, or change those to something different from the dimension, by the way.)
To sum up what's been discussed in the comments between OP and me, and so that it leaves the Unanswered category:
There seemed to be some confusion around the use of $\varepsilon$ both as the tool for the proof of $(R(x_n))_n$ being Cauchy (where it is fixed) and as the name of the variable for the definition of $(x_n)_n$ being Cauchy (where it is a "silent" variable). The critical point is that the latter can be given any value and name independent from the former's, it does not have to be our fixed $\varepsilon$, and this allows us to pick "$\varepsilon \leftarrow \displaystyle\frac{\varepsilon}{\|R\|}$".
A good idea at first, at least to make things clearer to the reader and/or yourself, is to begin the proof like this:
Let $(x_n)_n$ be a Cauchy sequence in $X$. We want to prove that $(R(x_n))_n$ is Cauchy in $\mathbb{K}^d$, that is:
$$\forall \varepsilon > 0,\,\exists N(\varepsilon) \in \Bbb N\, : \, \forall n,m \in \Bbb N, \, n,m > N(\varepsilon) \Rightarrow \| R(x_n) - R(x_m) \|_\infty < \varepsilon$$
Let $\varepsilon > 0$ be fixed. Since $(x_n)_n$ is Cauchy, we have:
$$\forall \delta >0,\, \exists M(\delta) \in \Bbb N\, : \, \forall n,m \in \Bbb N, \, n,m > M(\delta) \Rightarrow \| x_n - x_m \| < \delta$$
Define $N(\varepsilon) := M\left(\displaystyle\frac{\varepsilon}{\|R\|}\right)$, which is possible because $\frac{\varepsilon}{\|R\|} > 0$. [insert proof of the rest]
Hopefully that cleared any doubt around the $\varepsilon$s.
