Closure of the range of a compact operator Let $X$ be an infinite-dimensional Banach space, and let $Y$ be a banach. Let $T$ be a compact operator from $X$ to $Y$, ie. if $(x_n)$ is a sequence in $X$ then there is a subsequence s.t. $T(x_{n(k)})$ converges. We wish to show that 0 is in the closure of $\{Tx,||x||=1\}$.
I've been working on this problem for a while, and have been entirely unable to solve it without resorting to the spectral theorem, but I've been told there is an elementary solution involving only the fact that we can find a sequence of unit vectors $X$ s.t. that the distance between the elements is greater than or zero (ie. Riesz Lemma). 
 A: Suppose you have a continuous linear operator $A\colon X \to Y$ with $X$ and $Y$ as in the question, such that $0$ is not in the (norm) closure of $\{ Ax : \lVert x\rVert_X = 1\}$. Then
$$\delta := \inf \{\lVert Ax\rVert_Y : \lVert x\rVert_X = 1\} > 0.$$
Hence $A$ is injective, and $A$ is an open mapping to $\mathcal{R}(A)$; we have $\delta\lVert x\rVert_X \leqslant \lVert Ax\rVert_Y \leqslant \lVert A\rVert\cdot\lVert x\rVert_X$ for all $x\in X$, so $A^{-1} \colon \mathcal{R}(A) \to X$ is continuous, with $\lVert A^{-1}\rVert = \delta^{-1}$. Therefore $\mathcal{R}(A)$ is a Banach space (if $(Ax_n)$ is a Cauchy sequence in $\mathcal{R}(A)$, then $(x_n)$ is a Cauchy sequence in $X$, hence convergent, $x_n \to x_\ast$, and then $Ax_n \to Ax_\ast$ by the continuity of $A$). Since $A$ is injective, $\mathcal{R}(A)$ is infinite-dimensional. Hence $\overline{A(B_X)}$, which is a neighbourhood of $0$ in $\mathcal{R}(A)$ is not compact (since only finite-dimensional Hausdorff topological vector spaces are locally compact [Riesz]). Therefore $A$ is not compact.
