QUESTION#1 is why he required that the space Y is complete not only the range of the operators is complete?
QUESTION#2
In the proof of this theorem :
i take a cauchy sequence {Tn} from B(X,Y) and proved that for each x in X, {Tn(x)} is a cauchy sequence and since Y is complete this sequence {Tn(x)} has a limit which we can denote by $s_x$ and we can check that the elements {$s_x$ : x in X } satisfy the properties of a linear transformation so we can define T:X to Y by T(x)=$s_x$ and then check that T defines a bounded linear map
my question is in this part:is the following proof is true for proving that T defines a "bounded " linear map since Tn(x) convergent to T(x) then for every epsoln >0 there exist no in N :n>no implies ||Tn(x)-T(x)||< epsoln then can i say then that Tn-T is bounded and since Tn is bounded then T is bounded
I assume we're talking about normed linear spaces here, and (for convenience) real scalars.
1) If $Y$ is not complete, there is a Cauchy sequence $y_n$ in $Y$ that has no limit there. Take linear operators $T_n$ on $\mathbb R$ such that $T_n 1 = y_n$. Then these form a Cauchy sequence in $B(\mathbb R, Y)$ with no limit.
2) I'm not sure I understand your question, but pointwise convergence ($T_n x \to Tx$ for all $x$) does not in general imply convergence in norm. You really need uniform estimates.