If the cardinality of $B(X)$, the space of operators on $X$, is continuum, must $X$ be separable? Does there exits any non-separable Banach space $X$ such that the size (cardinal number) of $B(X)$, bdd linear operators on $X$, is just of the continuum? 
 A: An example of a non-separable Banach space $X$ with $|B(X)|=\mathfrak c$ is  any non-separable Banach space $X$ whose dual $X^*$ is $w^*$-separable and has cardinality $|X^*|=\mathfrak c$. 
This follows from the observation that the map $B(X)\to B(X^*)$, $T\mapsto T^*$, is injective and hence for a countable $w^*$-dense set $D$ in $X^*$ we have  $$|B(X)|\le |B(X^*)|\le |(X^*)^{D}|=|X^*|^\omega=\mathfrak c^\omega=\mathfrak c.$$
A ZFC-example of a non-separable Banach space $X$ whose dual space $X^*$ is $w^*$-separable and has cardinality $|X^*|=\mathfrak c$ is the Banach space $X=C(K)$ of continuous functions on the Alexandroff two-arrow space $K$. 
The $w^*$-separability of the dual space $X^*$ was proved by Corson, see Theorem 12.43 in this book. The equality $|X^*|=\mathfrak c$ can be seen analyzing the structure of (probability) measures on the compact space $K$.
A: Assume Martin's axiom and the negation of CH. Then $2^{\omega_1}=\mathfrak c$. Let $X=\ell_2(\omega_1)$. Every operator on $X$ is determined by its values on a dense set of cardinality $\omega_1$, hence there are at most $$|\ell_2(\omega_1)|^{\omega_1} = (\omega_1^\omega)^{\omega_1}\leqslant (2^\omega)^{\omega_1}=2^{\omega_1}=\mathfrak c$$ operators on $X$. Consequently, $|B(\ell_2(\omega_1))|=\mathfrak{c}$.
