Image of sphere is compact?

Question

Let $S$ be the unit sphere of a Banach space $X$, $K:X\to Y$ a compact operator. So why is $K(S)$ not surely compact in $Y$? We just know that it is totally bounded.

Answer 1

Because asserting that the operator $K$ is compact simply means (by definition) that $K(S)$ is relatively compact.

Answer 2

Consider $l^2$ and define $T(e_n)={1\over n}e_n$ it is compact since it is the limit of the operators $T_n$ such that $T_n(e_i)=0, i>n, T_n(e_i)={1\over i}e_i, i\leq n$. The image of the sphere does not contain $0$, but $0=lim_nT(e_n)$.