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Let $A$ be any nom-empty set, and let $\mathrm{B}(A)$ denote the set of all the bounded real or complex-valued functions with set $A$ as their domain, and let $d \colon \mathrm{B}(A) \times \mathrm{B}(A) \to \mathbb{R}$ be defined as $$d(x,y) \colon= \sup \left\{ \ \left\vert x(t) - y(t) \right\vert \ \colon \ t \in A \ \right\} \ \mbox{ for all } x, y \in \mathrm{B}(A).$$ Then this $d$ is a metric on $\mathrm{B}(A)$.

Is this metric space complete?

My effort:

Let $\left( x_n \right)_{n \in \mathbb{N}}$ be a Cauchy sequence in $\mathrm{B}(A)$. Then, given $\varepsilon > 0$, we can find a natural number $N$ such that $$ d\left( x_m, x_n \right) < \frac{\varepsilon}{2} \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N, n > N.$$ So, for each $t \in A$, we have $$ (1) \ \ \ \left\vert x_m(t) - x_n(t) \right\vert < \frac{\varepsilon}{2} \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N, n > N,$$ which implies that the sequence $\left( x_n(t) \right)_{n \in \mathbb{N}}$ is a Cauchy and therefore convergent sequence of real or complex numbers; let $x(t)$ denote the limit of this sequence.

In this way, we have a real or complex-valued function $x$ which is defined on set $A$.

Now in (1), we let $m \to \infty$, keeping $n$ fixed, and obtain $$ (2) \ \ \ \left\vert x_n(t) - x(t) \right\vert \leq \frac{\varepsilon}{2} \ \mbox{ for any natural number } n \mbox{ such that } n > N \mbox{ and for each } t \in A,$$ and so, for each $t \in A$, we have $$\left\vert x(t) \right\vert \leq \left\vert x_{N+1}(t) - x(t) \right\vert + \left\vert x_{N+1}(t) \right\vert \leq \frac{\varepsilon}{2} + \sup \left\{ \ \left\vert x_{N+1}(s) \right\vert \ \colon \ s \in A \ \right\} < +\infty.$$ Thus $x \in \mathrm{B}(A)$.

Finally, from (2) above, we can conclude that $$d \left( x_n, x \right) = \sup \left\{ \ \left\vert x_n(t) - x(t) \right\vert \ \colon \ t \in A \ \right\} < \varepsilon \ \mbox{ for any natural number } n \mbox{ such that } n > N,$$ showing that $\left(x_n \right)_{n \to \mathbb{N}}$ converges in $\mathrm{B}(A)$ to the point $x$.

Hence $\mathrm{B}(A)$ is complete.

Is there any problem with this proof?

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No. There is no problem with your proof. Well done !

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