There exist infinite-dimensional spaces X, Y and a compact surjective operator T : X → Y . I'm trying to understand this proof that shows that there exist infinite-dimensional spaces X, Y and a compact surjective operator
T : X → Y.
Let $K$ = {$(x_n) ∈ l^{2} : |x^n|\leq{1/n}$} (we recall that $l^2$ is the space of all real or complex sequences $(x_n)$ satisfying $\sum\limits_{n}|x^n|^2 <∞$ with the norm $||(x_n)|| = (\sum_n|x_n|^2)^{1/2})$ and let $X = \bigcup_n nK$ with the norm defined as the Minkowski functional of K ( $\mu_K(x)=inf$ {$t>0: t^{-1}x\in{K}$} ).
Let Y = X with the original $l^2$ norm and T : X → Y be the identity mapping. Then both the spaces X and Y are infinite-dimensional and T is a compact surjective operator.
What I understand is, The reason why this is posible, is because the space X is not complete. Here's my problem, I already proved $\mu_k$ is a norm in X, but I can't find a Cauchy sequence that doesn't converge in X (to prove X is not complete). Can anyone help me with this?
 A: It might be possible that $X$ is complete. Indeed, if $T : X \to Y$ is a bounded, linear, compact and surjective operator between infinite-dimensional Banach spaces, we can consider the completion $\tilde X$ of $X$ and extend $T$ to an operator $\tilde T : \tilde X \to Y$. It is still bounded, linear, compact and coercive, but now the domain $\tilde X$ is complete. However, the space $Y$ cannot be complete.
In your example, it is quite easy to check that $Y$ is a dense, proper subspace of $\ell^2$, i.e., it cannot be complete.
Edit: Your space $X$ is complete. Consider the linear operator $S : X \to \ell^\infty$ which maps a sequence $(x_n)$ to the sequence $(n \, x_n)$. It is easy to check that $S$ is well defined and an isometry. Further, $S$ is even surjective: for all $(y_n) \in \ell^\infty$, we have $(y_n/n) \in \ell^2$, thus, $(y_n/n) \in X$. Altogether, $S$ is an isomorphism and the completeness of $X$ follows since $\ell^\infty$ is complete.
A: The norm you have defined is simply given by $\mu_K(x)=sup{|x_n|/n}$. The sequence {1,1,...,1,0,0,...} is Cauchy in this norm but if it converges, then all the coefficients of the limit have to be 1.
