# Is it in general true that a space is not homeomorphic to the punctured version of this space?

For non-arbitary spaces we can discuss for such case, like how many components are there or other properties. But is it true for any space? It seems if we have a homeomorphism $$f$$ from $$S$$ to $$S' = S - \{p\}$$, $$f(p) = q$$, but since a space is homeomorphic to itself, there is some $$g(r)=q$$. Then there is no inverse if $$f$$ and $$g$$ coincide. However they don't have to and maybe $$f$$ is somehow the homeomorphism since I can't deduce more information.

If it is not true, a counter-example will be super helpful! Thank you.

• Even two different puncturings of a space needn’t be homeomorphic; it may matter which point is removed. Consider removing a point from a figure “X”. The result might have 1, 2, or 4 components, depending on which point is removed.
– MPW
Commented Dec 24, 2019 at 21:32

Another example: $$\mathbb C \setminus \mathbb Z$$. This is connected and homeomorphic to the punctured version of itself.
Any infinite discrete space is homeomorphic to itself minus any point. For example, the map $$n\mapsto (n+1)$$ is a homeomorphism $$\mathbb{N}\to \mathbb{N}\setminus\{0\}$$ (where $$\mathbb{N}$$ has the discrete topology).
The same example works if you give $$\mathbb{N}$$ the trivial topology or the Alexandrov topology (where open sets are upwards-closed sets).
$$\mathbb{Q}$$ is homeomorphic to its punctured version (all countable metric spaces without isolated points are). Same for the irrationals $$\mathbb{P}$$.