I have some struggles with this exercise.
Let $X$ and $Y$ be infinite dimensional Banach spaces. Show that no $T\in \mathcal{K}(X,Y)$ is onto. We assume that $T\in \mathcal{K}(X,Y)$, so we have that $T\in \mathcal{L}(X,Y)$. I am thinking of showing this by contradiction, so I assume that T is onto, then I have by the open mapping theorem that T is open. Since we have that X,Y are Banach spaces, we have that they are normed vector spaces, and we have that T is open, then we have that $B_Y(0,r)\subset T(B_X(0,1))$, then we'll have $\overline{B_Y(0,r)}\subset \overline{T(B_X(0,1))}$. Since we know that T is compact (since $T\in \mathcal{K}(X,Y)$) then we'll have that $\overline{T(B_X(0,1))}$ is compact hence $\overline{B_Y(0,r)}$ is compact. I am thinking that I will be done, if I can show that $\overline{B_Y(0,1)}$ is not compact, but I have trouble doing so.
