# Removing a closed subset from a surface gives a surface

In Ethan Bloch's "A First Course in Geometric Topology and Differential Geometry" the definition of a surface is the following:

A subset $$Q \subseteq \mathbb{R}^n$$ is called a surface if each point $$p \in Q$$ has an open neighborhood that is homeomorphic to $$intD^2$$.

Later you encounter an exercise that says that if we remove a closed subset of a surface it remains a surface. Intuitively it seems hilariously obvious, but I'm not sure if the following is complete/correct.

My attempt so far:

Let $$p \in Q \setminus F$$ where $$F \subseteq Q$$ is closed. There exists an open neighbourhood $$U \subseteq Q$$ such that $$p \in U$$ and there exists a homeomorphism $$h:U \rightarrow intD^2$$. Suppose that $$U \cap F \neq \emptyset$$. We consider the set $$A=U \cap(Q \setminus F)$$ which is an open neighbourhood of $$p$$ in $$Q \setminus F$$. Since $$h$$ is an open map we can find $$ε>0$$ so that $$B(h(p),ε) \subseteq h(A) \subseteq intD^2$$. Set $$V=h^{-1}(B(h(p),ε))$$. Now we have that $$p \in V \subseteq Q \setminus F$$ and $$V$$ is an open neighbourhood homeomorphic to an open ball in $$\mathbb{R}^2$$, hence to $$intD^2$$.

In case my attempt is not right it would be amazing if someone could help me fix it!

Your proof is correct. Your exercise is just a special case of a general assertion about topological manifolds, the latter being topological spaces which look locally like Euclidean space $$\mathbb{R}^n$$. As a first conceptual introduction you may have a look at https://en.wikipedia.org/wiki/Manifold.
In fact, if $$M$$ is an $$n$$-dimensional manifold, then each open subset of $$M$$ is again an $$n$$-dimensional manifold.