TITLE: Existence of injective compact operators
QUESTION [5 upvotes]: We know that if  $X$ is a separable Banach space, then for every infinite dimensional Banach space  $Y$, there exists an injective compact operator from $X$ to $Y$.
My query is for every  Banach space $X$ (need not be separable ) do there exist a Banach space  $Y$ and an injective compact operator  $T:X\to Y$?

REPLY [12 votes]: No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X = \ell_\infty^*$. Then you have no chance of building such operators.
Another possibility for counterexamples comes with non-separable reflexive spaces (or, more generally, WCG spaces), which contain non-separable weakly compact sets. Injective bounded linear operators are then homeomorphic embeddings of such sets with respect to the weak topology, so cannot be compact.