### Small note

One might have wondered why $B(X,Y)$ contains only linear bounded operators, and not linear operators of any and every kind. This has a very specific reason. Unless we fix $x\in X$, we cannot construct a cauchy sequence $\{T_1,T_2,\dots\}$ of linear operators. And we do not want to fix $x\in X$, as we want to define the limit $\lim\limits_{n\to\infty} T_n x$ for every $x\in X$.

Now if we allow ourselves to take all $x\in X$, knowing that $T_i (\alpha x)=|\alpha|T_i(x)$, we coud convert any sequence of linear operators into a divergent one. Hence, we need to construct sequences of the kind $\{T_i(x)/\|x\|\}$. This requires that we use bounded linear operators.