On changing from '<' to '$\le$' when taking limits (with norm $|\bullet|_p$) I'm reading Gouvêa's book on $p-$adic, and there's one problem that I don't think I really get it. Here's a proposition, and the problem attached to it. It's on page 57, 58 of the book.

Proposition 3.2.12
The image of $\mathbb{Q}$ under the inclusion $\mathbb{Q} \rightarrowtail \mathbb{Q}_p$ is a dense subset of $\mathbb{Q}_p$.
Proof
Take any element $\lambda \in \mathbb{Q}_p$, a Cauchy presentation $\{x_n \}$ of it, and any $\epsilon > 0$. We'll prove that there's a constant sequence that belongs to $B(\lambda, \epsilon)$. We now choose an $\epsilon'$ which is slightly less than $\epsilon$.
Since $\{ x_n \}$ is a Cauchy sequence, then there exists $N>0$, such that for all $n, m \ge N$, we'll have: $|x_m - x_n| < \epsilon'$. We claim that the constant sequence $\{ x_N \}$ will indeed satisfy the requirement.
For any $n \ge N$, we'll have $|x_n - x_N| < \epsilon'$.
Taking the limit as $n \to \infty$ yields $\lim |x_n - x_N| \color{red}{\le} \epsilon' < \epsilon$.
Which means that $\{ x_N \} \in B (\lambda , \epsilon)$.



Problem 87.
Why does < become $\le$ in the limit? Do we really need the business of decreasing $\epsilon$ slightly to $\epsilon'$?


Ok, honestly, I think that we don't (but the answer states that we do).
While it's true that I can find examples when the sign does change in taking limits, say: $x_n = p^n$, then of course $|x_n|_p = |p^n|_p = \dfrac{1}{p^n}>0, \forall n$, but $\lim |x_n|_p = 0$. I think I can do this because 0 is in fact, a limit point of the set $\{p^z | z\in \mathbb{Z} \}$.
However, I don't think there's any example of a Cauchy sequence, such that $|x_n|_p < \epsilon, \forall n$, but when we take the limit, it'll end up with $\lim |x_n|_p = \epsilon$.
Are there such sequences? Am I missing something here? Or is the book wrong?
Thank you guys very much,
And have a good day, :*
 A: You are correct and the book is wrong ... if the exercise was talking specifically about ${\bf Q}_p$.
In the reals, the reason we must replace $<$ with $\le$ in the limit is that the image of the archimedean absolute value is $[0,\infty)$, which has right-accumulation points (points that are accumulated to from the left and towards the right, a term I just made up on the spot). With any $\epsilon>0$ it is possible to obtain a sequence $(d_n)_{n=1}^\infty$ such that $\lim|d_n|=\epsilon$ but $|d_n|<\epsilon$ for all $n$.
In ${\bf Q}_p$ the image of the $p$-adic absolute value is $\{0\}\cup\{p^n:n\in{\bf Z}\}$; there are no right-accumulation points in this set so if $\lim\limits_{n\to\infty}|d_n|_p=\epsilon$ then either $\epsilon=0$ or $|d_n|_p$ is eventually the constant $\epsilon$; in the former case the strict inequality $|d_n|<\epsilon$ is impossible, so that case can be ruled out, and the other case implies that $|d_n|=\epsilon$ eventually, contradicting strict inequality.
In any extension (of valued fields) of ${\bf Q}_p$ with infinite ramification (by which I mean the image of the valuation / absolute value has nontrivial accumulation points), most prominently ${\bf C}_p$ (the $p$-adic metric completion of the algebraic closure of ${\bf Q}_p$), it actually is possible to obtain sequences $d_n$ such that $|d_n|_p\to\epsilon$ but $|d_n|_p<\epsilon$ for all $n$. For example $d_n=p^{1/n}$ and $\epsilon=1$.
Note that the question is just asking about the implication $|a_n|_p<c\implies \lim\limits_{n\to\infty}|a_n|_p<c$ in general for sequences; not specifically the sequences it gives for its purposes involving Cauchy sequences.
A: In general there are such sequences. Here’s one in $\Bbb Q$, with the usual absolute value: if $x_n=1-\frac1n$ for $n\in\Bbb Z^+$, then $|x_n|<1$ for each $n\in\Bbb Z^+$, but $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $1$, and $|1|=1$. The situation in $\Bbb Q_p$, however, is a bit different, because $|\cdot|_p$ has a set of values that is discrete except for $0$.
