Bijective Bounded Operator Extension: Where do the new elements go to?

Question

Given a dense, proper subset of complete spaces: $$X,Y\text{ both complete and }A\subsetneq\overline{A}=X$$

and an operator between them: $$T:A\to Y\text{ continuous, linear and bijective}$$

Now, extending the operator gives: $$T_E:X\to Y\text{ continuous, linear and bijective}$$

But where do the new elements go to since there is no "space" available?

Answer 1

Case 1:

If the inverse is also continuous then the subset is necessarily all of space: $$A\cong Y\Rightarrow A\text{ complete }\Rightarrow A\text{ closed}\Rightarrow A=\overline{A}=X$$

Case 2:

If the inverse is not continuous then the extension will be still surjective but no more injective: $$Y=\mathcal{R}(T)\subseteq\mathcal{R}(T_E)\subseteq Y\Rightarrow \mathcal{R}(T_E)=\mathcal{R}(T)=Y$$ $$\left(x\notin A, T_E(x)\in Y=\mathcal{R}(T)\right)\Rightarrow\left(x\neq a,T_E(x)=T(a)=T_E(a)\right)$$

Answer 2

The new elements I believe will be going to the same places as before, since if A is dense in X we have sequences $u_n\in A$ such that $u_n\to u\in \overline {A}\setminus A=X\setminus A $

So originally we have $Tu_n\to Tu\in Y $, now we have $T_Eu_n\to T_Eu=Tu $ except now $u\in X$