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?
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$\begingroup$ What is $B(X)$? $\endgroup$– Taras BanakhCommented Jun 7, 2018 at 18:51
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$\begingroup$ @TarasBanakh: Edited. $\endgroup$– ABBCommented Jun 7, 2018 at 19:03
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1$\begingroup$ @TomekKania The number of weakly compact operators is the number of weakly compact sets times the number of continuous operators mapping the unit ball to a given weakly compact set. The number of weakly compact sets should not exceed $|X^*|^\omega$ and if all weakly compact sets are metrizable (in the weak topology), then the number of continuous operators mapping the unit ball to a given weakly compact set also should not exceed $|X^*|^\omega$. So, for a Banach space $X$ with ``small'' weakly compact sets, the number of weakly compact operators should not exceed $|X^*|^\omega$. Right? $\endgroup$– Taras BanakhCommented Jun 9, 2018 at 6:54
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1$\begingroup$ I think it okay. Then this space will be ZFC counterexample impan.pl/pl/wydawnictwa/czasopisma-i-serie-wydawnicze/… $\endgroup$– Tomasz KaniaCommented Jun 9, 2018 at 7:59
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1$\begingroup$ The exists eveb simpler ZFC-example -- the Banach space $C(K)$ over the Alexandroff two-arrows space $K$, see my answer below. $\endgroup$– Taras BanakhCommented Jun 9, 2018 at 11:48
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2 Answers
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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 $K$.
answered Jun 9, 2018 at 11:15
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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, $|B(\ell_2(\omega_1))|=\mathfrak{c}$.
answered Jun 7, 2018 at 20:07
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$\begingroup$ Nice argument. What about under continuum hypothesis? $\endgroup$– ABBCommented Jun 8, 2018 at 5:44