Which topology is transferred by this specific metric on the rectangle $[0,\infty] \times [0 , \infty ]$?

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Let $\varphi : [0,\infty]^2 \to [0,1]^2$, $(t,x) \mapsto (\frac{t}{1+t} , \frac{x}{1+x})$. The inverse is given by $\varphi^{-1}: [0,1]^2 \to [0,\infty]^2$, $(t,x) \mapsto (\frac{s}{1-s} , \frac{x}{1-x})$. By this function $\varphi$ we can induce a metric $r$ from the euclidean metric on $[0,1]^2$.

Now consider the the set $\mathcal B$ of subsets $B$ of $[0,\infty]^2$ with

  1. $B$ is closed
  2. $\{\infty\}\times [0,\infty] \cup [0,\infty]\times \{\infty\} \subseteq B$
  3. $(t,x) \in B$ implies $(t,y) \in B$ for all $y>x$.

We can define a metric on $\mathcal B$ by $$r(C,D) = \max \{ \sup_{x\in C} \inf_{y\in D} r(x,y) , \sup_{x\in D} \inf_{y\in C} r(x,y) \}$$ EDIT: I found that this metric is called Hausdorff metric.

Now my question:

I found out that there is a bijection from $\mathcal B$ to the set $\underline {\mathbb b}$ of lower semi-continuous functions $b : [0,\infty] \to [0,\infty]$ with $b(\infty) = 0$ by $$b(t) = \inf\{x : (t,x ) \in B \} \\ B = \{(t,x) : x \geq b(t)\}$$

We can transfer the topology on $\mathcal B$ to $\underline {\mathbb b}$.

I would like to know which topology this is. For example, when does a sequence converge in $\underline {\mathbb b}$?

I tried to go step by step. For example I thought that for a sequence $B_n$ converging to $B$ at least one could show that the corresponding functions $b_n$ converge pointwise to $b$. But I have difficulties to work with this metric and came not really far.

EDIT: In the meantime I got the intuition, that $B_n \to B$ does not imply that the corresponding functions converge pointwise. For example, take a countable dense subset $D =\{d_1, \ldots \}$ of $[0,\infty]$. Then build a sequence of functions $b_n (t) = \infty$ for all $t\in [0,\infty]\setminus\{d_1 , \ldots , d_n\}$ and $b_n \equiv 1$ on $\{d_1 , \ldots , d_n\}$. Then the corresponding barrier should converge to $[1,\infty]\times [0, \infty ]$ but the functions clearly do not converge pointwise to $1$.

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The convergence I searched for is the so called $\Gamma$-Convergence