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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 $T:X\to Y$?

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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.

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    $\begingroup$ I was actually trying to check that it never exists if $X$ is non-separable. $\endgroup$
    – YCor
    Commented May 10, 2020 at 12:16
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    $\begingroup$ @YCor, take $T\colon \ell_\infty\to c_0$ given by $T (\xi_k)_{k=1}^\infty = (\xi_k / k)_{k=1}^\infty$. It is compact and injective. $\endgroup$
    – Tomasz Kania
    Commented May 10, 2020 at 12:17
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    $\begingroup$ Thanks, I can see why I was stuck! In any case this provides useful context. $\endgroup$
    – YCor
    Commented May 10, 2020 at 12:18
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    $\begingroup$ Perhaps it is worth pointing out that $X$ has the property if and only if $X^*$ is weak$^*$ separable, which is a somewhat weaker condition than that $X$ embeds isomorphically into $\ell_\infty$. $\endgroup$
    – Bill Johnson
    Commented May 10, 2020 at 13:59
  • $\begingroup$ @BillJohnson, with the first known example being your space $J\! L$ :-) $\endgroup$
    – Tomasz Kania
    Commented May 10, 2020 at 15:39

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