Set of bounded Cauchy sequences has empty interior. If $M$ is a complete metric space, prove that the set of bounded Cauchy sequences has empty interior in $\mathscr{B}(\mathbb{N};M$) with respect to sup metric, any tips would help.
In fact the original question asks to show that the set $\mathscr{D}$ of non Cauchy sequences is open and dense in $\mathscr{B}(\mathbb{N};M)$, I did the first part, but I'm struggling with density, so I was trying to use some equivalences, as the one above, but I'm not going anywhere.
First I tried to construct a open ball $B(g; \varepsilon)$, $g \in \mathscr{B}(\mathbb{N};M)$ and see that $B(g; \varepsilon) \cap \mathscr{D} \neq \emptyset$, but I couldn't think of a construction of such an sequence in this ball that satisfies what I want.
If $g \in \mathscr{D}$ is trivial that $B(g; \varepsilon) \cap \mathscr{D} \neq \emptyset$, but if $g \notin \mathscr{D}$ I don't how to proceed or if this way is easiest one.
And the last thing, I'm suspecting that the book I'm reading forgot to include the additional condition that $M$ don't have any isolated points, so I need to know if this is necessary to solve this question.
 A: Assume that $M$ is non-empty and has no isolated points. We want to show that non-Cauchy sequences are dense in the set of all bounded sequences. All we need to show is that we can approximate any Cauchy-sequence arbitrary well by non-Cauchy sequences.
Pick a Cauchy sequence $(x_n)_{n\in \mathbb{N}}\in \mathcal{B}(\mathbb{N};M).$ As $M$ is complete, our sequence admits a limit $x:=\lim_{n\rightarrow \infty} x_n\in M$. As $M$ has no isolated points, there exists for every $\varepsilon >0$ some $x\neq y_\varepsilon\in B(x, \varepsilon ).$ As $x$ is the limit, we have that for every $\varepsilon>0$ exists some $N_\varepsilon$ such that $d_M(x_n,x)<\varepsilon$ for $n\geq N_\varepsilon$. Now define
$$z_n^{(\varepsilon)} := \begin{cases}x_n,& n<N_\varepsilon, \\ y_\varepsilon,& n\geq N_\varepsilon, n \text{ even}, \\ x,& n\geq N_\varepsilon, n \text{ odd}. \end{cases}$$
The sequence $(z_n^{(\varepsilon)})_{n\in \mathbb{N}}$ is bounded, non-Cauchy and approximates $(x_n)_{n\in \mathbb{N}}$ as well as we like for $\varepsilon\rightarrow 0^+$. Indeed, we have
$$ d_{\mathcal{B}(\mathbb{N};M)} ((x_n)_{n\in \mathbb{N}},(z_n^{(\varepsilon)})_{n\in \mathbb{N}}) = \sup \{ \sup_{n< N_\varepsilon} d_M(x_n, z_n^{(\varepsilon)}), \sup_{n\geq N_\varepsilon} d_M (x_n, z_n^{(\varepsilon)})\} = \sup \{ 0, \sup_{n\geq N_\varepsilon} d_M (x_n, z_n^{(\varepsilon)})\}
\leq \sup_{n\geq N_\varepsilon} (d_M(x_n, x)+d_M(x_n, y_\varepsilon))<3\varepsilon.$$
The idea here is that the Cauchy sequence converges, so after a certain index $x_n\approx x$ and then in order to make it non-Cauchy, we replace the tail by an oscillation.
If we replace $M$ by some nontrivial Banach space, then the contruction is a bit easier. Pick your favourite bounded, non-Cauchy sequence $(y_n)_{n\in \mathbb{N}}$. Then the sequence $(x_n+\varepsilon y_n)_{n\in \mathbb{N}}$ is bounded, non-Cauchy and we have
$$ d_{\mathcal{B}(\mathbb{N}, M)}((x_n+\varepsilon y_n)_{n\in \mathbb{N}}, (x_n)_{n\in \mathbb{N}})=\varepsilon \Vert (y_n)_{n\in \mathbb{N}}\Vert_{\mathcal{B}(\mathbb{N}, M)} \rightarrow 0$$
for $\varepsilon\rightarrow 0^+$. This was my starting point and then tried to remove the dependence on the additive structure.
