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!

Thank you in advance!

  • 1
    $\begingroup$ This looks all correct to me. $\endgroup$ Nov 2 '18 at 12:41
  • $\begingroup$ Thank you for your answer! $\endgroup$ Nov 2 '18 at 17:05

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.


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