Fix a manifold $M$ along with an embedding of $M$ into $\R^N$ and set $X=\Ran(M)\times \R_{\geqslant 0}$. The goal of this post is to show that every $(P,t)\in X$ has an open neighborhood that contains no points of the type $(Q,d(Q_i,Q_j))$, for some $i\neq j$. The collection of all such elements of $X$ is called the

Following Lurie, given a collection of open sets $\{U_i\}_{i=1}^k$ in $M$, set

\[\Ran(\{U_i\}_{i=1}^k) = \left\{P\in \Ran(M)\ :\ P\subset \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i\right\}.\]

The topology on $\Ran(M)$ is the smallest topology for which every $\Ran(\{U_i\}_{i=1}^k)$ is open, for any $\{U_i\}_{i=1}^k$, for any $k$. The topology on the product $X$ is the product topology.

Fix $(P,t)\in X$ not in the singularity set of $X$, with $P=(P_1,\dots,P_k)$, for $1\leqslant k\leqslant n$. Set

\[\mu = \min\left\{t,\min_{1\leqslant i<j\leqslant k}\left\{|t-d(P_i,P_j)|\right\}\right\},\]

with distance $d$ being Euclidean distance in $\R^N$. The quantity $\mu$ should be thought of as the upper bound on how "far" we may move from $(P,t)$ without hitting the singularity set.

\[U=\Ran\left(\{B(P_i,\alpha/2)\}_{i=1}^k\right) \times \left(t-\beta,t+\beta\right)\]

is an open neighborhood of $(P,t)$ in $X$ and does not contain any points of the singularity set of $X$.

If $t=0$, then having $[0,\beta)$ as the second component of $U$, with $\alpha+\beta=\min_{i,j}d(P_i,P_j)$ works as the open neighborhood of $(P,t)$. The balls $B(x,r)$ are $N$-dimensional in $\R^N$. The proof is mostly applications of the triangle inequality.

*singularity set*of $X$, as the Vietoris-Rips complex at $Q$ with such a radius changes at such elements.Following Lurie, given a collection of open sets $\{U_i\}_{i=1}^k$ in $M$, set

\[\Ran(\{U_i\}_{i=1}^k) = \left\{P\in \Ran(M)\ :\ P\subset \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i\right\}.\]

The topology on $\Ran(M)$ is the smallest topology for which every $\Ran(\{U_i\}_{i=1}^k)$ is open, for any $\{U_i\}_{i=1}^k$, for any $k$. The topology on the product $X$ is the product topology.

**Remark:**Note that the Ran space $\Ran(M)$ by itself can be split up into the pieces $\Ran^k(M)$, with "singularities" viewed as when a point splits into two (or more) points, or two (or more) combine into one. Then every element of $\Ran(M)$ is on the edge of the singularity set, as any neighborhood of a single point on the manifold contains two points on the manifold.Fix $(P,t)\in X$ not in the singularity set of $X$, with $P=(P_1,\dots,P_k)$, for $1\leqslant k\leqslant n$. Set

\[\mu = \min\left\{t,\min_{1\leqslant i<j\leqslant k}\left\{|t-d(P_i,P_j)|\right\}\right\},\]

with distance $d$ being Euclidean distance in $\R^N$. The quantity $\mu$ should be thought of as the upper bound on how "far" we may move from $(P,t)$ without hitting the singularity set.

**Proposition:**Let $(P,t)$ be as above and $t,\alpha,\beta>0$ such that $\alpha+\beta=\mu$. Then\[U=\Ran\left(\{B(P_i,\alpha/2)\}_{i=1}^k\right) \times \left(t-\beta,t+\beta\right)\]

is an open neighborhood of $(P,t)$ in $X$ and does not contain any points of the singularity set of $X$.

If $t=0$, then having $[0,\beta)$ as the second component of $U$, with $\alpha+\beta=\min_{i,j}d(P_i,P_j)$ works as the open neighborhood of $(P,t)$. The balls $B(x,r)$ are $N$-dimensional in $\R^N$. The proof is mostly applications of the triangle inequality.

*Proof:*By construction we have that $U$ is open in $X$ and that it contains $(P,t)$. For $(Q,s)\in U$ any other element, we have three cases. We will show that the distance between any two $Q_a,Q_b\in Q$ is never $s$. Fix distinct indices $\ell,m\in \{1,\dots,k\}$.-
__Case 1:__$Q_a,Q_b\in B(P_\ell,\alpha/2)$. The situation looks as in the diagram below. Observe that $d(Q_a,Q_b)\leqslant d((Q_a,P_\ell)+d(Q_b,P_\ell) <\alpha = \mu-\beta \leqslant t-\beta$. Hence $d(Q_a,Q_b)<s$. __Case 2:__$Q_a\in B(P_\ell,\alpha/2), Q_b\in B(P_m,\alpha/2), d(P_\ell,P_m)>t$. The situation looks as in the diagram below. Observe that $d(P_\ell,P_m)\leqslant d(P_\ell,Q_b)+d(P_m,Q_b)\leqslant d(P_\ell,Q_a)+d(Q_a,Q_b)+d(P_m,Q_b) < \alpha+d(Q_a,Q_b)$. Since $d(P_\ell,P_m)>t$, the definition of $\mu$ gives us that $\mu \leqslant d(P_\ell,P_m)-t$, so combining this with the previous inequality, we get $d(Q_a,Q_b) > d(P_\ell,P_m)-\alpha\geqslant \mu+t-(\mu-\beta)=t+\beta$. Hence $d(Q_a,Q_b)>s$.__Case 3:__$Q_a\in B(P_\ell,\alpha/2), Q_b\in B(P_m,\alpha/2), d(P_\ell,P_m)<t$. The situation looks as in the diagram below. Observe that $d(Q_a,Q_b)\leqslant d(P_m,Q_b) + d(P_m,Q_a) \leqslant d(P_\ell,Q_a)+d(P_\ell,P_m)+d(P_m,Q_a)<\alpha+d(P_\ell,P_m)$. Since $d(P_\ell,P_m)<t$, the definition of $\mu$ gives us that $\mu\leqslant t-d(P_\ell,P_m)$, so combining this with the previous inequality, we get $d(Q_a,Q_b)<\mu-\beta+t-\mu = t-\beta$. Hence $d(Q_a,Q_b)<s$. $\square$

*References:*Lurie (Higher Algebra, Section 5.5.1)