**Definition:** A filtered space $X$ of formal dimension $n$ is *locally cone-like* if for all $i$, $0 \le i \le n$, and for each $x \in X^i - X^{i-1} = X\_i$ there is an open neighborhood $U$ of $x$ in $X\_i$, a neighborhood $N$ of $x$ in $X$, a compact filtered space $L$, and a homeomorphism $h:U \times cL\rightarrow N$ such that $h(U \times c(L^k)) = X^{i + k + 1}\cap N.$


**Note:** $cL$ denotes the open cone on $L$.


My confusion arises when considering the suspension $\Sigma S^1$. Give it the filtration $\{v,w\} \subset \Sigma S^1$, where $v$ and $w$ are the singular points. Setting $i=0$, we check $X\_0 = \{v,w\}\setminus\emptyset = \{v,w\}$. As far as I can tell, the first condition already fails because there is no open set $U \subset \{v,w\}$. The only work-around I see is if we're allowed to let $U = \emptyset$. Shouldn't $\Sigma S^1$ be locally cone-like?



Confusion about locally cone-like spaces
Just to confirm your self-answer in the comments: That's right. In this case if you're looking for a neighborhood of $v$ in $X\_0=X^0=\{v,w\}$, then $U=\{v\}$ does it and the cone neighborhood $N$ in $\Sigma S^1$ is $\{v\}\times cS^1$.


