Cauchy sequences in a linear normed space form a subspace of the space of bounded sequence

Question

$\mathbb{X}$ is a linear normed space.
The set $b(\mathbb{X})$ of sequences $(x_n)_{n\geq1}$ with values in $\mathbb{X}$, that are bounded, and $$||(x_n)_{n\geq1}||_{*}=\sup_{n\geq 1}{||x_n||}<\infty$$ We check easily that $b(\mathbb{X})$ is a normed linear space.
I want to prove that Cauchy sequences in $\mathbb{X}$ form a subspace say $b_c(\mathbb{X})$, of $b(\mathbb{X})$.
I know that every Cauchy sequence is bounded, so $b_c(\mathbb{X})$ is a subset of $b(\mathbb{X})$. If $x_n\in b(\mathbb{X})$, then $x_n+y_n \in b_c(\mathbb{X})$ and $\alpha x_n \in b_c(\mathbb{X})$. Therefore, $b_c(\mathbb{X})$ is a algebraic subspace of $b(\mathbb{X})$.
But I don't know how to prove $b_c(\mathbb{X})$ is closed.

Answer 1

It follows from the triangle inequality $|a+b|\leq|a|+|b|$. Assume that you have a sequence $(x_n)_{n\in\mathbb{N}}$ of Cauchy sequences $x_n = (x_n^k)_{k\in\mathbb{N}}$ in $b(X)$ that converges to some limit sequence $x\in b(X)$. Then you have to show that $x$ is again a Cauchy sequence:

Let $\varepsilon > 0$, then there exists a $N\in\mathbb{N}$ with $||x-x_n||_\infty \leq \frac{\varepsilon}{3}$ for all $ n \geq N$. Also, there exists an $M\in\mathbb{N}$, so that $|x_N^n-x_N^m| \leq \frac{\varepsilon}{3}$ for all $n,m \geq M$.

Thus, for any $m,n >M$, we have (using the triangle inequality) $$|x^m-x^n| \leq \underbrace{|x^m-x_N^m|}_{\leq \|x-x_N\|_\infty\leq \frac{\epsilon}{3}} + \underbrace{|x_N^m-x_N^n|}_{\leq\frac{\varepsilon}{3}} + \underbrace{|x^n-x_N^n|}_{\leq \|x-x_N\|_\infty\leq \frac{\epsilon}{3}} \leq \varepsilon,$$

Where the first and last terms are smaller than $\frac{\varepsilon}{3}$ due to $x$ being the limit of the sequence $(x_n)_n$ (with respect to the norm $\|\cdot\|_\infty$ of $b(X)$), and the middle term is smaller than $\frac{\varepsilon}{3}$ because $(x_N^k)_{k\in\mathbb{N}}$ is a Cauchy-sequence.