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If $M$ is a metric space then the set of Cauchy sequences is closed in $\mathcal{B}(\mathbb{N},M)$ (set of bounded functions $x : \mathbb{N} \rightarrow M$).

My attempt:

Let $A= \lbrace x \in \mathcal{B}(\mathbb{N},M) | \space x_n \space \text{is Cauchy} \rbrace$.

I tried to show that $\mathcal{B}(\mathbb{N},M) - A = \lbrace x:\mathbb{N} \rightarrow M | \space x_n \space \text{is not Cauchy} \rbrace$ is open.

If $x \in \mathcal{B}(\mathbb{N},M) - A$ then $x_n$ is not a Cauchy sequence, so

$ \forall N \in \mathbb{N}, \exists \epsilon>0$ so that $m,n >N \rightarrow d(x_n,x_m)> \epsilon $

Now, fix some $N \in \mathbb{N}$ and consider the ball $B(x,\frac{\epsilon}{4})= \lbrace y \in \mathcal{B}(\mathbb{N},M) | \quad\|x-y\| < \frac{\epsilon}{4} \rbrace$

Let $y \in B(x, \frac{\epsilon}{4}) \Rightarrow \|x-y\|< \frac{\epsilon}{4} \Rightarrow \sup_{n \in \mathbb{N}} d(x_n,y_n) < \frac{\epsilon}{4}$

Then, if $m,n >N$ we have $d(x_n,y_n)<\frac{\epsilon}{4}$

Now, if $m,n>N$ we have that

$d(y_n,y_m) > d(x_n,y_m) - d(x_n,y_n)>d(x_n,x_m)-d(x_m,y_m)-d(x_n,y_n)>\epsilon -\frac{2\epsilon}{4}=\frac{\epsilon}{2}$

Then, $y_n$ is not a Cauchy sequence. So, $B(x,\frac{\epsilon}{4}) \subset \mathcal{B}(\mathbb{N},M)-A$

Then, every point in $\mathcal{B}(\mathbb{N},M)-A$ is a interior point, follows that $\mathcal{B}(\mathbb{N},M)-A$ is open and $A$ is closed.

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It's just fine. That's how I would have done it. Two minor suggestions:

  1. You should have stated which distance you're using in $\mathcal{B}(\mathbb{N},M)$.
  2. Plase type $A\setminus B$ and not $A-B$. Sometimes, the later option is ambiguous.
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There is just one error. You have not correctly negated the statement "$x$ is a Cauchy sequence". You have written:

If $x \in \mathcal{B}(\mathbb{N},M) - A$ then $x_n$ is not a Cauchy sequence, so

$ \forall N \in \mathbb{N}, \exists \epsilon>0$ so that $m,n >N \rightarrow d(x_n,x_m)> \epsilon $.

It should instead be:

If $x \in \mathcal{B}(\mathbb{N},M) - A$, then $\{ x_n \}_{n \in \mathbb{N}}$ is not a Cauchy sequence, so

$\exists\ \epsilon > 0$ such that $\forall\ N \in \mathbb{N}$, there exist $m,n > N$ such that $d(x_n,x_m) \geq \epsilon$.

So, the part:

Now, if $m,n>N$ we have that ...

needs to be modified slightly, into:

Now, choose $m,n > N$ such that $d(x_n,x_m) \geq \epsilon$. Then, for this value of $m,n \in \mathbb{N}$ we have that ...

Apart from this, the proof is fine.

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