Generalizing Gromov Hausdorff distance using Vietoris topology

There are two notions of convergence of a sequence of metric space. One is by the Gromov Hausdorff distance for compact metric spaces, another one is the pointed Gromov Hausdorff convergence for pointed (possibly non-compact) metric spaces. We here suggest a third one, for sequences of complete (possibly non-compact) metric space.

Let $Y$ be a metric space, and $CL(Y)$ be the set of closed subsets of $Y$. We can equip $CL(Y)$ with the Vietoris topology. Let $\mathcal O\subset CL(Y)$ be a basic open set. Denote $\mathcal N(Y,\mathcal O)$ to be the set of all metric space $Z$ (up to isometry) such that there exists an isometric embedding $f$ of $Z$ into the interior of $Y$ with $f(Z)\in\mathcal O$. If I did not make mistake, it seems that we can use the collection of all possible $\mathcal N(Y,\mathcal O)$ (we run over all possible O and Y) as a basis, to define a topology $\tau$ on the set of all isometry classes of complete metric space. As a result, given a sequence of complete metric spaces, we can define the notion of convergence using $\tau$.

Is this definition already know? Can someone please suggest a reference? Moreover, is there a theorem for this topology $\tau$ analog to the Gromov compactness theorem?

By Taras’s comment: Given a sequence of complete separable metric space such that when embedded in the Urysohn universal space, their images converge under the Vietoris topology. Then does this garuntee that the sequence converges under $\tau$, and this limit space is isometric to the one in the Urysohn universal space?

Edit: I modified the definition to require an embedding of Z into the interior of Y, to allow convergence of a sequence of metric from outward, eg an decreasing sequence of set under inclusion.

• Your topology depends on $Y$ (?) it seems to be interesting only in case if any metric space admits a distance preserving map into $Y$ (?) if so it remind qute a bit GH-topology. Now do you want non-isometric metric space to have positive distance --- it seems that your metric does not have this property (?) – Anton Petrunin Feb 16 '18 at 4:29
• @Anton The topology does not depend on Y, because for the collection of all N(Y,O) we run over all possible O and Y. – A. Chu Feb 16 '18 at 4:46
• What about nonisometric spaces with zero distance from each other? – Anton Petrunin Feb 16 '18 at 5:44
• Maybe to fulfill all the construction in the universal Urysohn space? Will this give the same convergence for separable complete metric spaces? – Taras Banakh Feb 16 '18 at 6:38
• @ Anton, I suppose you mean nonisometric spaces that are indistinguishable under the topology $\tau$. I cannot think of such an example yet, do you have one? – A. Chu Feb 16 '18 at 6:41