I'm reading a paper and I have a doubt about about a part of it.
In the paper it is defined a topology over $\mathcal{F}(X)$ the set of all closed sets of a polish space $X$, which is named Wijsman topology.
To define this topology, we choose $\rho$ a complete compatible metric on $X$, $\{\alpha_j\}_{j=1}^\infty$ a dense sequence on $X$ and we define $T: \mathcal{F}(X) \longrightarrow \mathbb{R}^\mathbb{N}$ such that $$T(F)=\left\{\rho(\alpha_j, F) \right\}_{j=1}^\infty$$
Wijsman topology on $\mathcal{F}(X)$ is then defined as the initial topology induced by $T$ when we consider the product topology on $\mathbb{R}^\mathbb{N}$.
After the definition, it is proven that $\mathcal{F}(X)$ equipped with this topology is polish and the set $$\left\{ (x,F) \in X \times \mathcal{F}(X): x \in F \right\}$$ is closed in $X \times \mathcal{F}(X)$.
Now, it seems that base on this last afirmation, the authors conclude that Wijsman topology doesn't depend on the dense sequence we take to define $T$, but I don't really see why this is the case.
I upload an image of the paragraph where all of this is discussed in the paper (condition (iii) is that the set $\left\{ (x,F) \in X \times \mathcal{F}(X): x \in F \right\}$ is closed).
Any ideas?
Thanks a lot.
It seems (I don't have Baer's book on hyperspaces, but it's mentioned here) that the Wijsman topology has an alternative definition (not depending on a dense subset) as follows: we identify a closed set $F$ by the function $d_A: X \to \Bbb R, d_A(x)=d(x,A)$ and then $\{d_A\mid A \in \mathcal{F}(X)\}$ is a subspace of $C_p(X)$ (the continuous real-valued functions in the pointwise (i.e. product) topology and we give $\mathcal{F}(X)$ the topology from that identification. This already seems very similar to the definition in your paper.