$X-\{x\}$ is homeomorphic to $X-\{y\}$ for any $x\ne y$? It's easy to see that a topological space $X$ is homeomorphic to itself via the identity map. Is $X-\{x\}$ homeomorphic to $X-\{y\}$ for any $x\ne y$? If yes, what are homeomorphisms can be defined here? If no, what are conditions on $X$ so that the statement holds true?
Thank you.
 A: I don't think this is true. Consider $X = (a,b) \cup \{c\}$ as a subspace of $\mathbb{R}$ where $a < b < c$. let $a < y < b$.Then $X-\{c\}$ is connected while $X - \{y\}$ is not. Thus there can be no homeomorphism between these sets.
A: This is true if $X$ is a connected topological manifold. Indeed, according to this post, for any two points $x,y\in X$ there exists a homeomorphism $\phi$ satisfying $\phi(x) =y$. This homeomorphism restricts to a homeomorphism from $X\setminus \{x\}$ to $X\setminus\{ y\}$.
This is not true in general for connected topological spaces and even for manifolds with boundary. The space $X=[0,1]$ is a counterexample. Removing any interior point disconnects $X$ while removing a boundary point does not. 
Added later:
More generally, the property you mentioned holds for topological spaces that are homogeneous (with respect to their homeomorphism group). A topological space $X$ is said to be homogeneous if for any $x,y \in X$ there exists a homeomorphism of $X$ sending $x$ to $y$. As the argument given above shows, this implies that $X \setminus \{x \}$ and $X \setminus\{ y\}$ are homeomorphic for any $x,y$. As remarked above, connected topological manifolds are important examples of homogeneous spaces. There are a lot of other examples, such as topological groups, the Cantor set and any set with the discrete topology or the trivial topology.
A: It's almost never true. For example, consider $[0,1]$. Deleting any middle point yields a disconnected set, while deleting either endpoint yields a connected set.
To make $X - \{x\}$ always homeomorphic to $X - \{y\}$, you need a very strong homogeneity condition on $X$ - intuitively, you need every point to "look" the same. But it can't even be just a local property, because this property you're talking about can't be preserved by disjoint unions.
