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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 & \{<Tx,y>\ :\ 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?

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Actually $F$ 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 $x,y$. – Robert Israel Nov 14 '12 at 5:06

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It is a homeomorphism onto its image (not onto the whole product) because it maps each basic weak-operator open set to a product-topology open set in the image.

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If I take a basic open set on $\mathcal{B}(H)$ containing $T_0$ of the form $\{T\in\mathcal{B}(H)\ : \ <(T-T_0)x_j,y_j> < \epsilon \ (j=1,... , m) \}$, can I characterize its image under F ? – Felipe Pérez Nov 14 '12 at 3:16
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As I commented, you want to restrict to the unit ball of ${\cal B}(H)$. The image under $F$ is those $w \in \prod_{x,y} {\mathbb D}_{x,y}$ such that $w_{x_j, y_j} - \langle T_0 x_j, y_j \rangle < \epsilon$ for $j = 1 \ldots m$ and $w_{ax+by,z} = a w_{x,z} + bw_{y,z}$ and $w_{x,ay+bz} = a w_{x,y} + bw_{x,z}$ for all scalars $a,b$ and all $x,y,z \in H$. – Robert Israel Nov 14 '12 at 5:14
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For the same reason. If numbers $\langle T_jx,y\rangle$ converge to $\langle Tx,y\rangle$, then this means precisely that $T_j\to T$ in the WOT.

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Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets... – Robert Israel Nov 14 '12 at 5:00
I never said sequences, did I? ;) – Martin Argerami Nov 14 '12 at 5:06

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