I'm reading Richard Kadison's book about operator algebras, and in the demonstration that the unit ball is compact in weak-operator topology, the author defines a function from the set of bounded operators on a Hilbert space $H$, to a product of disks: \begin{align*}F:\mathcal{B}(H)\rightarrow&\prod_{x,y\in H}\mathbb{D}_{x,y}\\ T\rightarrow & \{\langle Tx,y\rangle\ :\ x,y\in H\} \end{align*} If we set the product topology on $\prod_{x,y\in H}\mathbb{D}_{x,y}$, the function above is continuous because of the topology of $\mathcal{B}(H)$ is induced by functions of that type, but I can't see why this function is a homeomorphism. Why the inverse is also continuous?
Compactness of unit ball in weak-operator topology.
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analysis
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1Actually $F$ should not be defined on all of $\mathcal{B}(H)$, just on the unit ball, and ${\mathbb D}_{x,y}$ should be the closed disk centred at $0$ with radius $\|x\| \|y\|$. If $T$ is not in the unit ball, $\langle Tx, y\rangle$ will be outside that disk for some $x,y$. – 2012-11-14