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Let $A$ be the open unit disc in $\mathbb{R}^2$ and $B$ be the closed unit disc in $\mathbb{R}^2$.

The toplogical boundary of $A$ and $B$ is $S^1$. This I understand.

The manifold boundary of $A$ is $\emptyset$ and the manifold boundary for $B$ is $S^1$. This I don't understand. Can someone briefly explain?

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2 Answers 2

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You need to review your definition of a manifold with boundary v.s. manifold without boundary.

In the case of A, for every point $p \in A$ you can find a neighbourhood $U$ such that $p \in U$ and a diffeomorphism $\phi: U \to V \subset \mathbb{R}^2$ (the natural inclusion), where $V$ is open in $\mathbb{R}^2$.

In the case of B, for every point $p \in S^1 \subset B$ you cannot find a neighborhood $U$ of $p$ such that $U$ is diffeomorphic (even homeomorphic) to some open subset $V$ of $\mathbb{R}^2$, but you can find one diffeomorphic to an open subset of some half-space $\mathbb{R}^2_{\lambda \geq c}$ making it a manifold with boundary.

NOTE: When you say that the topological boundary of $A$ is $S^1$ you are thinking of A as a subset of $\mathbb{R}^2$ and when you think of A as a manifold you think it as the whole space.

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  • $\begingroup$ Sorry for replying to this old post, but I'm struggling to see how the topological and manifold boundaries are related. Isn't it simply that in the first case, $\partial A$ is the $\mathbb R^2$-boundary (i.e. the boundary with respect to the topology on $\mathbb R^2$) and in the second case, it is the $A$-boundary (i.e. the boundary with respect to the subspace topology on $A$ induced by $\mathbb R^2$)? $\endgroup$
    – 0xbadf00d
    Jul 6, 2020 at 10:13
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    $\begingroup$ @0xbadf00d No problem. Yes, $\partial A$ is the $\mathbb{R}^2$-boundary. Notice that under the subspace topology, any subset $A \subset X$ of a topological space has empty (topological) boundary. Take $x \in A$ and an open set $U \subset X$ such that $x \in U$. Then $U \cap A$ is an open set in the subspace topology and $U \cap A \subset A$, hence $x$ is an interior point of $A$. So, for example, the closed unit disk $B$ has empty topological boundary under the subspace topology but a non-empty boundary both as a subset of $\mathbb{R}^2$ and as a manifold. $\endgroup$
    – Luis Vera
    Jul 7, 2020 at 13:36
  • $\begingroup$ Thank you for your reply. You're right and I guess that the situation doesn't change if we take the topological boundary with respect to the topology on $\mathbb R^2$ instead, right? One thing I've noticed is that the manifold boundary is always a subset of the manifold itself. So, I guess this gives rise to another counterexample, right? $\endgroup$
    – 0xbadf00d
    Jul 8, 2020 at 17:04
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In a topological manifold $M$ (possibly with boundary), each point $p \in M$ must have an open neighborhood $p \in U$ which is homeomorphic either to an open set in $\mathbb{R}^n$ or to an open set in $\mathbb{H}^n := \{ (x_1, \ldots, x_n) \in \mathbb{R}^n \, | \, x_n \geq 0 \}$. The set $\partial \mathbb{H} := \{ (x_1, \ldots, x_{n-1}, 0) \, | \, x_1, \ldots, x_{n-1} \in \mathbb{R} \}$ is called the boundary of $\mathbb{H}$.

Let $p \in M$ and let $\varphi \colon U \rightarrow V$ be a homeomorphic between an open neighborhood of $p \in U$ in $M$ and an open neighborhood in $\mathbb{R}^n$ or $\mathbb{H}^n$. If $V \subseteq \mathbb{H}^n$ and $\varphi(p) \in \partial \mathbb{H} \cap V$, we say that $p$ is a boundary point of $M$. If $V \subseteq \mathbb{R}^n$ or $V \subseteq \mathbb{H}^n$ and $\varphi(p) \notin \partial \mathbb{H}^n$, we say that $p$ is an interiour point. It is a non-trivial result that the set of interior points and the set of boundary points is disjoint.

Consider for example the closed unit disk $B = B(0,1) \subseteq \mathbb{R}^2$. Using polar coordinates, you can easily construct a homeomorphism from an open neighborhood of each point $x \in B$ with $|x| = 1$ to an open neighborhood of $\mathbb{H}^2$ mapping $x$ to $\partial \mathbb{H}^2$. For $x \in B$ with $|x| < 1$, the open ball $\{ x \, | \, |x| < 1 \}$ is a neighborhood that is an open subset of $\mathbb{R}^2$. Thus, the boundary of the closed unit disk as a manifold is the same as the topological boundary. Since the open unit disk is already an open subset of $\mathbb{R}^2$, the boundary of it considered as a manifold is empty.

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  • $\begingroup$ Sorry for replying to this old post, but I got a question on it. I'm used to the following definition: $M$ is a $n$-dimensional submanifold of $\mathbb R^N$ if for all $p\in M$ there is a homeomorphism $\psi$ from an open neighborhood $\Omega_1\subseteq\mathbb R^N$ of $p$ onto an open subset $\Omega_2$ of $\mathbb R^N$ such that with $U:=\Omega_1\cap M$ either $$\psi(U)=\Omega_2\cap(\mathbb R^n\times\{0\})\tag1$$ ($0\in\mathbb R^{N-n}$) or $$\psi(U)=\Omega_2\cap(\mathbb H^n\times\{0\})\text{ and }\psi(p)\in\partial\mathbb H^n\times\{0\}\tag2.$$ $\endgroup$
    – 0xbadf00d
    Jul 7, 2020 at 5:02
  • $\begingroup$ The corresponding chart is $\phi:=\pi\circ\left.\psi\right|_U$, where $\pi$ is the projection of $\mathbb R^N$ onto the first $n$-components. If $(1)$ holds, then $p$ is an interior point; if $(2)$ holds, it is a boundary point. This is slightly different from your definition. I've got two problems with this: (a) This definition seems to be unsound, since if $(2)$ is satisfied, $(1)$ is satisfied as well and so they are not disjoint. So, is this definition broken? $\endgroup$
    – 0xbadf00d
    Jul 7, 2020 at 5:02
  • $\begingroup$ (b) I've read that we can assume that if $p$ is an interior point, then $\phi_k(p)>0$. Why? By the definition above we should only know that if $(2)$ doesn't hold, then $\phi_k(p)\ne0$. I've asked for that here: math.stackexchange.com/q/3748169/47771. $\endgroup$
    – 0xbadf00d
    Jul 7, 2020 at 5:02
  • $\begingroup$ Forget (a), it doesn't make sense what I wrote. $\endgroup$
    – 0xbadf00d
    Jul 7, 2020 at 5:55
  • $\begingroup$ What is $k$ in $\phi_k(p) > 0$? If $\mathbb{H}^n = \{ (x_1, \dots, x_n) \, | \, x_n \geq 0 \}$ then in the first case, you can always replace $\Omega_1$ with a smaller open set such that $\psi(U)$ is bounded and then compose $\phi$ with a translation to make $\phi_n(p) > 0$. Note that according to your definition, if $2$ doesn't hold, there's nothing that guarantees that $\phi_n(p) \neq 0$. That's not a problem with the definition as you can always compose everything with a translation to make $\phi_n(p) > 0$. $\endgroup$
    – levap
    Jul 7, 2020 at 12:47

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