Is there a (what is the) intrinsic definition of boundary?

Question

It is asked to show that the closed disk $\overline{D}^2=\{(x,y)\in \Bbb{R}^2:x^2+y^2\leq 1\}$ (with the topology induced from $\Bbb{R}^2$) is not a regular surface.

It seems obvious that we have a problem already at the topological level (not needing to go into differentiability): $\overline{D}^2$ has nonempty boundary.

It is pretty pictoric that we can't have a homeomorphism between an open set from $\Bbb{R}^2$ and an open set containing a point from the boundary, but is there a way to prove this rigorously and by simple means?

This question has already been done here and here but, in the first, the answer is not very satisfactory and, in the second, there is no answer at all, only a comment.

Maybe the following is at stake:

What is the intrinsic definition of boundary?

Because if the definition of the boundary of $X$ is $\partial X:=\overline{X}\backslash \overset{\circ}{X}$, if $X$ is the universe set then always $\partial X=\overline{X}\backslash \overset{\circ}{X}=X\backslash X=\emptyset$... which is unpleasant...

I would suggest that a point $p\in X$ is said to be at the boundary if, there is no closed path $\gamma:S^1\to X$ "surrounding $p$". I mean, there is no closed path, determining a simply connected region in $X$ such that $p$ is this region, but not over the trace $\gamma(S^1)$. Is this correct? How to make this more precise? What rigourously would mean "to determine a region", etc?

EDIT: I was thinking also if the following definition works (for the bidimensional case): a point $p\in X$ is said to be in the boundary of the space $X$ if there is a simply connected neighborhood $N$ of $p$ such that $N-\{p\}$ is also simply connected.

Answer 1

First of all, I think it's appropriate to stress that the notion of boundary for (subspaces of) topological spaces is different of that for topological/smooth manifolds. A canonical example (as you already discussed in the comments) is the open ball, which has empty boundary as a manifold, but has the sphere as boundary as subspace of the euclidean space. Another easy example is that the topological boundary, because the definition involves an ambient space, is aways empty for the ambient itself (as you already mentioned too).

For the manifold case, the notion is clearly defined intrinsically, and your intuition with the homotopy groups is on spot, but it's easier to work with the relative homology groups instead (the geometric intuition is the same):

$x \in \text{Int}(M)$ if and only if $H_n(M,M-x;\mathbb{Z}) \approx \mathbb{Z}$;

$x \in \partial M$ if and only if $H_n(M,M-x;\mathbb{Z}) \approx 0$.

but it only works because topological manifolds are very well-behaved spaces.

The definition of the topological boundary makes sense only for subspaces, but its preserved by homeomorphisms: if $f:X\to Y$ is a homeomorphism, then $f(\partial(A))=\partial(f(A))$ for every $A\subset X$, so it's natural. It will depend now on what you understand by intrinsic.

Answer 2

I am just trying to answer my own question by collecting the several ideas in the comments. Thank to you all guys!

Proposition: Open disks and half disks (open at the arc and closed at the diameter) are not homeomorphic.

Proof. (Not very rigorous yet...): Suppose there is a homeomorphism $h:D_1\to D_2$ between a disk and a half disk. Consider a point $p$ in the diameter of the half disk $D_2$. Then $h|_{D_1-\{h^{-1}(p)\}}:{D_1-\{h^{-1}(p)}\}\to D_2-\{p\}$ is still a homeomorphism. Since ${D_1-\{h^{-1}(p)\}}$ is not simply connected, but $D_2-\{p\}$ is, this is a contradiction. $\blacksquare$

This allow us to give the following definition in dimension 2:

Definition: Let $X$ be a topological space. We say that $X$ is a topological surface if, for every point $p\in X$, there is an open neighborhood $N\ni p$ and a homeomorphism $h:N\to D$ satisfying one of the conditions:

1. $D$ is an open disk;
2. $D$ is a half disk and $h(p)$ belongs to the diameter of it.

By the proposition above, a point $p\in X$ cannot have both of these properties simultaneously. So a topological surface is a topological space such that every point is either of type-1 or the type-2. Those of type-1 are called the interior points of the space $X$ and those of type-2 are the boundary points of the space $X$.

For greater dimensions, we can use open $n$-balls and $n$-half balls (we use higher homology groups to show that a open $n$-ball is not homeomorphic to an $n$-half ball).

Answer 3

Here is an elementary proof using only properties of the fundamental group. Let me use $\overline H$ for the closed upper half plane, and $\overset{\circ}D$ for the open 2-disc.

Given a 2-manifold with boundary $X$ and a point $x \in X$, the following hold:

Interior point: There exists a neighborhood $U \subset X$ of $x$ and a homeomorphism $U \to \overset{\circ}D$ if and only if $x$ has a neighborhood basis in $X$ of the form $U_1 \supset U_2 \supset \cdots$ such that there exist homeomorphisms $U_i \to \overset{\circ}D$, and such that each of the inclusion maps $U_i - \{x\} \to U_{i-1}-\{x\}$ is homotopically nontrivial (proof of the "only if" direction: define $U_i$ to be the inverse image under the homeomorphism $U \to \overset{\circ}D$ of the open disc of radius $1/i$).

Boundary point: There exists a neighborhood $V \subset X$ of $x$ and a homeomorphism $(V,x) \to (\overline H,0)$ if and only if $x$ has a neighborhood basis in $X$ of the form $V_1 \subset V_2 \subset \cdots$ for which there exist homeomorphisms $(V_i,x) \to (\overline H,0)$ (proof of the "only if" direction: define $V_j$ to be the intersection with $\overline H$ of the open disc of radius $1/j$). Note that each set $V_j - \{x\}$ is simply connected.

From the definition of a 2-manifold with boundary, for each $x \in X$ we know that at least one of these two properties holds.

So we just have to prove that it is impossible for both of them to hold.

Suppose that it is possible.

Let $U_{i_1}$ be one of the interior point neighborhood basis elements.

Let $V_j \subset U_{i_1}$ be a boundary point neighborhood basis element.

Let $U_{i_2} \subset V_j$ be an interior point neighborhood basis element.

Consider the composition of the two injection maps $$U_{i_2}-\{x\} \to V_j - \{x\} \to U_{i_1}-\{x\} $$ This composition is the injection map $U_{i_2}-\{x\} \to U_{i_1}-\{x\}$ and is therefore homotopically nontrivial. But this composition factors through the simply connected set $V_j - \{x\}$ and so is homotopically trivial. Contradiction.

In my topology class, this is one of the first few examples that I use to emphasize that the concept of a "topological invariant" is more than just a property that is invariant under homeomorphisms: a topological invariant is a functor.