Collar neighborhood theorem for manifold with corners I was reading this wonderful sequence of posts:
nlab: manifold with boundary
and nlab: collar neighbourhood theorem
and I couldn't help but wonder.  Is there an extension of the Collar neighborhood Theorem for manifolds with corners; in the sense that if $X$ is a $d$-dimensional manifold with corners then the set $X_0$ of points $x \in X$ with no neighborhood homeomorphic to $\mathbb{R}^d$ has a neighborhood $U_{X_0}$ which is homeomorphic to $X_0\times [0,1)$ such that this homomorphism maps $X_0$ to $X_0\times \{0\}$?
Edit: @MoisheKohan both solved and refined my initial formulation.  Here is the updated version which I initially hoped to express:
Does there exist such a $U_{X_0}$ which is isomorphic to $X_0\times [0,1)$ as manifolds with corners?
 A: This is an extended comment.
The way your question is stated, the answer is
"yes," of course. Each (topological) manifold with corners is also a manifold with boundary. Your subset $X_0$ is just the boundary. Now, the claim follows from the existence of a collar neighborhood of the boundary.
There is a more refined version of your question where instead of treating $X_0$ as a topological space, you treat it as a  manifold with corners. Then $U=U_{X_0}$ is also a stratified manifold with corners, where each stratum is a manifold with corners, and so is $X_0\times [0,1)$. The refined question is then if there exists $U$ such that the manifolds with corners $U$ and $X_0\times [0,1)$ are isomorphic in the category of stratified topological manifolds with corners. I am quite sure that the answer to this one is also positive and is proven by imitating the standard argument for the existence of a collar neighborhood. But I did not think about the details since it is unclear if this is what you want to know. With the naive (product) stratification of the product space, such an isomorphism of stratified manifolds fails. I will have to think if there is a reasonable modification of the product stratification under which one still has an isomorphism.
