I have a functional space as below. Let A be a non empty space and $$ B(A) := \{ f : A \rightarrow \mathbb {R} \lvert \, \lVert f \rVert_\infty : = \sup_{a \in A} \, \lvert f(a) \rvert < \infty \} $$
I prove like this Let $x_n $ be a cauchy sequence in $ B(A)$. The for $ \epsilon >0 \,, \exists N \in \mathbb N , \forall \, m,r $ $$ \lVert f_{m} - f_r \rVert < \epsilon $$ $$ \sup_{a \in A} \, \lvert f^{(m)}(a) -f^{(r)}(a) \rvert < \epsilon $$ $$ \lvert f^{(m)}(a) -f^{(r)}(a) \rvert < \epsilon \, \forall a \in A $$ This is a cauchy sequence in $\mathbb R$. $\mathbb R$ is complete space. The sequence converges for every $a\in A$. Therefore lets define $f : A \rightarrow \mathbb R$ such that as $ m \rightarrow \infty \implies f^m (a) \rightarrow f(a) \implies f(a) \in B(A)$ Then from $$ \lVert f_{m} - f_r \rVert < \epsilon $$ we have as $ r \rightarrow \infty $ $$ \lVert f_{m} - f \rVert < \epsilon $$ $$ \sup_{a \in A} \, \lvert f^{(m)}(a) -f(a) \rvert < \epsilon $$ $$ \lvert f^{(m)}(a) -f(a) \rvert < \epsilon \, \forall a \in A $$ Therefore $B(A)$ is compelte space . Hence it is banach space. Is this proof correct or there are some flaws.