Deleting a subset from a topological manifold

Question

Consider an $n$-topological manifold $M$. We remove a subset $A$ from $M$. Are there cases where $M-A$ is no longer a topological manifold. In case we suppose that $M-A$ is still a manifold, what should be the dimension of $M-A$, is it exactly $n$ or some other integer $\leq n$ ? thank you !!

Answer 1

Here's the complete answer to part of your question: If $M$ is a topological manifold of dimension $n$ and $A\subseteq M$, then $M\smallsetminus A$ is a topological manifold of dimension $n$ (in the subspace topology) if and only if $A$ is closed in $M$.

One direction is easy to prove: If $A$ is closed, then $U=M\smallsetminus A$ is an open subset of $M$, and an open subset of an $n$-manifold is easily shown to be an $n$-manifold.

Conversely, suppose $M\smallsetminus A$ is an $n$-manifold in the subspace topology. We'll show that $U = M\smallsetminus A$ is open in $M$. Given $p\in U$, there exist a neighborhood $V$ of $p$ in $U$ and a homeomorphism $\phi\colon V\to \widehat V$ to some open subset $\widehat V\subseteq\mathbb R^n$. If $\iota\colon U\hookrightarrow M$ is the inclusion map, then $\iota\circ\phi^{-1}$ is an injective continuous map from $\widehat V$ into $M$, and it follows from invariance of domain that $\iota(\widehat V) = V$ is open in $M$. Thus each point of $U$ has an open neighborhood in $M$, so $U$ is open.

The question of when $M\smallsetminus A$ is a manifold of some dimension less than $n$ is much more complicated, and I doubt that there's any simple necessary and sufficient condition.