2024-10-04

Last time (2024-09-27) we constructed a pseudometric which gave a condition on $ϵ$ such that $H({\text{Nrv}}_{St}(\mathcal{U}))$ and $H({\text{Nrv}}_{St}(\mathcal{V}))$ are $ϵ$-interleaved. We now seek a condition for which $H({\text{Nrv}}_{St}(\mathcal{U}))$ is $q$-tame. Obviously if $\mathcal{U}$ is a finite cover, then we are done. So we work in the case that $\mathcal{U}$ is infinite. It suffices to show that it $ϵ$-interleaves with a finite dimensional space. Making use of our current result we just need to construct an $ϵ$-simplicial correspondence to a finite space.

In Persistence stability for geometric complexes they show that if $X$ is totally bounded then $H(\mathbb{R}\text{ips}(X))$ is $q$-tame by showing it $ϵ$-interleaves with an $ϵ/2$-sample for all $ϵ$. We try the same strategy.

Our goal is given a cover $\mathcal{U}$ constructed by the MAPPER, construct a correspondence $C:\mathcal{U}\rightrightarrows \mathcal{V}$ to a finite cover $\mathcal{V}$ that satisfies

$$|d_{St}(\{U_i\mid i\in \sigma \})-d_{St}(\{V_i\mid i\in \tau \})|<\frac{1}{2}ϵ,\forall (\sigma ,\tau )\in C.$$

Assume that $\mathcal{U}$ is a cover in a totally bounded space. Then there is a finite collection of open balls $\mathcal{V}=\{V_i{\}}_{i=1}^n$ of radius $ϵ/2$ such that

$$\bigcup _{U_i\in \mathcal{U}}{U}_i\subseteq \bigcup _{i=1}^n{V}_i.$$

Define the map

$$C=\{(U,V)\in \mathcal{U}\times \mathcal{V}\mid d_{St}(U,V)<\frac{1}{2}ϵ\}.$$

Question: Is this map well-defined? I’m fairly certain it is not.

Notice then that

$$\begin{array}{rl}|d_{St}(\{U_i\mid i\in \sigma \})-d_{St}(\{V_j\mid j\in \tau \})|& =\left|\frac{\mu (\bigcup V_j)}{\mu (\bigcap V_j)}-\frac{\mu (\bigcup U_i)}{\mu (\bigcap U_i)}\right|\\ ???& \end{array}$$

---

Assume that $\mathcal{U}$ is a cover in a totally bounded space $(X,d_X)$. Let $F$ be an $ϵ/2$-sample of $X$, i.e., for all $x\in X$ there is some $f\in F$ such that $d_X(x,f)<ϵ/2$. For each cover element $U\in \mathcal{U}$ construct the set

$$V=\{y\in F\mid \exists x\in U\text{ such that }{d}_X(x,y)<ϵ/2\}.$$

We then define the map $C:\mathcal{U}\to \mathcal{V}$ given by $C(U)=V$. Since $X$ is totally bounded, $F$ is finite. So $\mathcal{V}$ is also finite.

We want to show that for all $(\sigma ,\tau )\in C$

$$|d_{St}(\{U_i\mid i\in \sigma \})-d_{St}(\{V_i\mid i\in \tau \})|<\frac{1}{2}ϵ.$$

No clue how to do this…

Also if we used the Lebesgue measure in $\mathcal{V}$ land, then all the measures would be zero.

---

Assume that $\mathcal{U}$ is a cover of a compact space. Then $\mathcal{U}$ has a finite subcover $\mathcal{V}\subseteq \mathcal{U}$. Define the map $C:\mathcal{U}\to \mathcal{V}$ defined by

$$C(U)=\text{arg}\underset{V\in \mathcal{V}}{\min }\{d_{St}(U,V)\},$$

i.e., the cover element $V$ closest to $U$ in terms of Steinhaus distance. Note that if $U\in \mathcal{V}$, then $C(U)=U$.

Then for all $(\sigma ,\tau )\in C$,

$$\begin{array}{rl}|d_{St}(\{U_i\mid i\in \sigma \}-d_{St}(\{V_i\mid i\in \tau \})|& =|d_{St}(\{U_i\mid i\in \sigma \}-d_{St}(\{U_i\mid i\in \tau \})|.\end{array}$$

Claim: $\tau$ is a face of $\sigma$.

In such a case, we get that this absolute difference is the birth time of $\sigma$ minus the birth time of $\tau$. However, it is not clear that this difference can be made arbitrarily small.