[proofplan] We work directly from the definition of a smooth manifold with boundary: every point admits a chart modelled on an open subset of the closed half-space $\mathbb{H}^n = \{x \in \mathbb{R}^n : x_n \ge 0\}$. The strategy has three stages. First, we recall the dichotomy between interior and boundary points and verify that it is intrinsic, so that $\partial M$ is well-defined. Second, given any $p \in \partial M$, we restrict a boundary chart $(U, \varphi)$ at $p$ to $U \cap \partial M$ and project away the last coordinate; the result is a chart on $\partial M$ whose image is an open subset of $\mathbb{R}^{n-1}$ — **not** of a half-space. Third, we conclude that every point of $\partial M$ is an interior point of $\partial M$, so $\partial M$ has empty boundary. [/proofplan]

[step:Fix notation and recall the definition of a smooth manifold with boundary] Let \begin{align*} \mathbb{H}^n &= \{x = (x_1, \dots, x_n) \in \mathbb{R}^n : x_n \ge 0\} \end{align*} be the closed upper half-space, equipped with the [subspace topology](/page/Subspace%20Topology) from $\mathbb{R}^n$. Its topological boundary in $\mathbb{R}^n$ is \begin{align*} \partial \mathbb{H}^n &= \{x \in \mathbb{R}^n : x_n = 0\}, \end{align*} which we identify with $\mathbb{R}^{n-1}$ via the map $(x_1, \dots, x_{n-1}, 0) \mapsto (x_1, \dots, x_{n-1})$. The topological interior of $\mathbb{H}^n$ is $\operatorname{Int} \mathbb{H}^n = \{x : x_n > 0\}$. By hypothesis, $M$ is a smooth $n$-manifold with boundary: $M$ is a second-countable Hausdorff topological space equipped with a maximal smooth atlas $\mathcal{A} = \{(U_\alpha, \varphi_\alpha)\}$, where each chart map \begin{align*} \varphi_\alpha &: U_\alpha \to \varphi_\alpha(U_\alpha) \end{align*} is a homeomorphism from an [open set](/page/Open%20Set) $U_\alpha \subseteq M$ onto an open subset $\varphi_\alpha(U_\alpha)$ of $\mathbb{H}^n$, and the transition maps $\varphi_\beta \circ \varphi_\alpha^{-1}$ are smooth in the sense of extending smoothly to an open neighbourhood in $\mathbb{R}^n$ where defined. A point $p \in M$ is called an **interior point** if there exists a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$ and $\varphi(p) \in \operatorname{Int} \mathbb{H}^n$. It is called a **boundary point** if there exists a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$ and $\varphi(p) \in \partial \mathbb{H}^n$. The boundary $\partial M$ is the set of boundary points. [/step]

[step:Verify that the interior/boundary dichotomy is intrinsic] We must check that no point of $M$ is simultaneously an interior point and a boundary point: otherwise $\partial M$ would not be well-defined as a subset on which to discuss further boundaries. [claim:No point of $M$ is both an interior and a boundary point] [proof] Suppose $p \in M$ has two charts $(U, \varphi), (V, \psi) \in \mathcal{A}$ with $p \in U \cap V$, $\varphi(p) \in \operatorname{Int} \mathbb{H}^n$, and $\psi(p) \in \partial \mathbb{H}^n$. We derive a contradiction. Set $W = U \cap V$, an open neighbourhood of $p$ in $M$. The transition map \begin{align*} \tau &= \psi \circ \varphi^{-1} : \varphi(W) \to \psi(W) \end{align*} is a homeomorphism between open subsets of $\mathbb{H}^n$, and by the smooth-manifold-with-boundary structure it extends to a smooth diffeomorphism $\tilde\tau: \widetilde{\varphi(W)} \to \widetilde{\psi(W)}$ between open subsets of $\mathbb{R}^n$ (the tildes denote the open extensions). Now $\varphi(p) \in \operatorname{Int} \mathbb{H}^n$ means there exists $r > 0$ such that the open ball $B(\varphi(p), r) \subseteq \varphi(W) \cap \operatorname{Int} \mathbb{H}^n$. Thus $\tilde\tau$ restricted to $B(\varphi(p), r)$ is a smooth diffeomorphism onto an open subset of $\mathbb{R}^n$ containing $\psi(p)$. Since the differential $D\tilde\tau_{\varphi(p)}$ is invertible, the image $\tilde\tau(B(\varphi(p), r))$ is open in $\mathbb{R}^n$ by the [Inverse Function Theorem](/theorems/30). On the other hand, $\tilde\tau(B(\varphi(p), r)) = \tau(B(\varphi(p), r)) \subseteq \psi(W) \subseteq \mathbb{H}^n$, and this image is a neighbourhood of $\psi(p)$ in $\mathbb{R}^n$. But $\psi(p) \in \partial \mathbb{H}^n$ has $\psi(p)_n = 0$, so any neighbourhood of $\psi(p)$ in $\mathbb{R}^n$ contains points with negative last coordinate, contradicting containment in $\mathbb{H}^n$. This contradiction shows that no point of $M$ is simultaneously an interior and a boundary point. [/proof] [/claim] Consequently, $\partial M$ is a well-defined subset of $M$, and a point $p$ belongs to $\partial M$ if and only if $\varphi(p) \in \partial \mathbb{H}^n$ for **every** chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$. [/step]

[step:Construct boundary charts on $\partial M$ by projecting out the last coordinate]Let $p \in \partial M$ be arbitrary. Choose a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$. Define the projection \begin{align*} \pi: \mathbb{R}^n &\to \mathbb{R}^{n-1} \\ (x_1, \dots, x_n) &\mapsto (x_1, \dots, x_{n-1}). \end{align*} Set $U' := U \cap \partial M$, and define the candidate chart map \begin{align*} \varphi': U' &\to \mathbb{R}^{n-1} \\ q &\mapsto \pi(\varphi(q)). \end{align*} We verify the following properties of $(U', \varphi')$. **(i) $U'$ is open in $\partial M$.** This is immediate: $U' = U \cap \partial M$ is the intersection with $\partial M$ of an [open set](/page/Open%20Set) $U \subseteq M$, hence open in the [subspace topology](/page/Subspace%20Topology) of $\partial M$. **(ii) $\varphi(U') = \varphi(U) \cap \partial \mathbb{H}^n$.** Since the boundary classification is intrinsic by Step 2, we have for $q \in U$: \begin{align*} q \in \partial M \iff \varphi(q) \in \partial \mathbb{H}^n. \end{align*} Therefore $\varphi(U') = \{\varphi(q) : q \in U, \varphi(q) \in \partial \mathbb{H}^n\} = \varphi(U) \cap \partial \mathbb{H}^n$. **(iii) $\varphi'(U')$ is open in $\mathbb{R}^{n-1}$.** Since $\varphi(U)$ is open in $\mathbb{H}^n$, by definition of the [subspace topology](/page/Subspace%20Topology) there exists an [open set](/page/Open%20Set) $\Omega \subseteq \mathbb{R}^n$ with $\varphi(U) = \Omega \cap \mathbb{H}^n$. Then \begin{align*} \varphi(U) \cap \partial \mathbb{H}^n = \Omega \cap \partial \mathbb{H}^n, \end{align*} and the projection $\pi$ restricts to a homeomorphism $\partial \mathbb{H}^n \to \mathbb{R}^{n-1}$ (its inverse is $(y_1, \dots, y_{n-1}) \mapsto (y_1, \dots, y_{n-1}, 0)$). Hence $\varphi'(U') = \pi(\Omega \cap \partial \mathbb{H}^n)$ is open in $\mathbb{R}^{n-1}$, being the image of an [open set](/page/Open%20Set) under a homeomorphism (between $\partial \mathbb{H}^n$ and $\mathbb{R}^{n-1}$). **(iv) $\varphi': U' \to \varphi'(U')$ is a homeomorphism.** The map $\varphi'$ is the composition of $\varphi|_{U'}: U' \to \varphi(U) \cap \partial \mathbb{H}^n$ with the projection $\pi: \partial \mathbb{H}^n \to \mathbb{R}^{n-1}$. The first map is a homeomorphism because $\varphi$ is a homeomorphism on $U$ and $U' = \varphi^{-1}(\partial \mathbb{H}^n) \cap U$; the second is a homeomorphism as noted in (iii). The composition of homeomorphisms is a homeomorphism.[/step]

[guided]We are constructing, for each chart $(U, \varphi)$ of $M$ containing a boundary point, an induced chart on $\partial M$. The recipe is: intersect the chart domain with $\partial M$, then "throw away" the last coordinate, which is forced to be $0$ on the boundary anyway. Let $p \in \partial M$ be arbitrary, and pick a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$. Define \begin{align*} \pi: \mathbb{R}^n &\to \mathbb{R}^{n-1}, \quad (x_1, \dots, x_n) \mapsto (x_1, \dots, x_{n-1}), \end{align*} and set $U' = U \cap \partial M$, $\varphi' = \pi \circ \varphi|_{U'}$. We check four things about $(U', \varphi')$: **(i) $U'$ open in $\partial M$.** Definitionally, $\partial M$ carries the [subspace topology](/page/Subspace%20Topology) from $M$, and $U' = U \cap \partial M$ is the intersection of $\partial M$ with an [open set](/page/Open%20Set) of $M$. Done. **(ii) The image $\varphi(U')$ is exactly $\varphi(U) \cap \partial \mathbb{H}^n$.** This is where the intrinsic-ness of the boundary (Step 2) is consumed. Without that result, a point $q \in U$ with $\varphi(q) \in \partial \mathbb{H}^n$ might fail to be in $\partial M$, or vice versa. But Step 2 guarantees the equivalence, so the chart map cleanly identifies $U \cap \partial M$ with $\varphi(U) \cap \partial \mathbb{H}^n$. **(iii) The image $\varphi'(U')$ is open in $\mathbb{R}^{n-1}$.** The point here is that $\varphi(U)$ is open in $\mathbb{H}^n$ — but we want openness in $\mathbb{R}^{n-1}$, not in some half-space of $\mathbb{R}^{n-1}$. Write $\varphi(U) = \Omega \cap \mathbb{H}^n$ for some open $\Omega \subseteq \mathbb{R}^n$ ([subspace topology](/page/Subspace%20Topology)). Slicing by $\{x_n = 0\}$: \begin{align*} \varphi(U) \cap \partial \mathbb{H}^n = \Omega \cap \{x_n = 0\}. \end{align*} Under the identification $\partial \mathbb{H}^n \cong \mathbb{R}^{n-1}$ via $\pi$, an open slice of $\mathbb{R}^n$ at $x_n = 0$ becomes an [open set](/page/Open%20Set) in $\mathbb{R}^{n-1}$ — this is just the statement that the inclusion $\mathbb{R}^{n-1} \hookrightarrow \mathbb{R}^n$ (as the $x_n = 0$ hyperplane) is a topological embedding. So $\varphi'(U')$ is open in $\mathbb{R}^{n-1}$. **Note that no half-space appears**: this is the entire content of the theorem at the local level. **(iv) $\varphi'$ is a homeomorphism onto its image.** The map factors as $U' \xrightarrow{\varphi|_{U'}} \varphi(U) \cap \partial \mathbb{H}^n \xrightarrow{\pi} \varphi'(U')$. The first arrow is a restriction of a homeomorphism to a subset and its image, hence a homeomorphism. The second is a coordinate forgetting whose inverse is "insert a $0$ in the last slot", manifestly continuous. Both factors are homeomorphisms, so the composition is.[/guided]

[step:Verify the smooth atlas property and conclude $\partial(\partial M) = \varnothing$]The collection \begin{align*} \mathcal{A}' &= \{(U \cap \partial M, \pi \circ \varphi|_{U \cap \partial M}) : (U, \varphi) \in \mathcal{A} \text{ with } U \cap \partial M \ne \varnothing\} \end{align*} covers $\partial M$: indeed, by the definition of $\partial M$, every $p \in \partial M$ lies in some boundary chart $(U, \varphi) \in \mathcal{A}$, and then $p \in U \cap \partial M$ with image $\varphi'(U') \subseteq \mathbb{R}^{n-1}$ open by Step 3. The transition maps in $\mathcal{A}'$ are smooth: given two charts $(U, \varphi), (V, \psi) \in \mathcal{A}$ with $U \cap V \cap \partial M \ne \varnothing$, the transition $\psi' \circ (\varphi')^{-1}$ on $\varphi'(U \cap V \cap \partial M)$ is the restriction of the smooth transition $\psi \circ \varphi^{-1}$ (extended smoothly to a neighbourhood in $\mathbb{R}^n$) to the slice $\{x_n = 0\}$, post-composed with $\pi$ and pre-composed with the inclusion $(y_1, \dots, y_{n-1}) \mapsto (y_1, \dots, y_{n-1}, 0)$. Composition of smooth maps is smooth. Thus $\mathcal{A}'$ is a smooth atlas on $\partial M$ in which **every chart has image in $\mathbb{R}^{n-1}$ proper** — not in a half-space $\mathbb{H}^{n-1}$. Consequently, every point of $\partial M$ admits a chart $(U', \varphi')$ with $\varphi'(p) \in \varphi'(U') \subseteq \mathbb{R}^{n-1}$, and $\varphi'(U')$ is open in $\mathbb{R}^{n-1}$. To finish, we apply the boundary/interior classification of Step 2 to the smooth $(n-1)$-manifold-with-boundary structure on $\partial M$. Viewing $\mathbb{R}^{n-1}$ as $\operatorname{Int} \mathbb{H}^{n-1}$ (a subset of $\mathbb{H}^{n-1}$ disjoint from its boundary $\partial \mathbb{H}^{n-1}$), every chart in $\mathcal{A}'$ sends its domain into $\operatorname{Int} \mathbb{H}^{n-1}$. By the intrinsic-classification claim of Step 2 (applied now to $\partial M$ in place of $M$), every point of $\partial M$ is an interior point of $\partial M$, and no point of $\partial M$ is a boundary point of $\partial M$. Therefore \begin{align*} \partial(\partial M) &= \varnothing, \end{align*} which is the claim.[/step]

[guided]We assemble the local construction of Step 3 into a global smooth atlas on $\partial M$, then read off the conclusion. **Atlas property.** The collection \begin{align*} \mathcal{A}' = \{(U \cap \partial M, \pi \circ \varphi|_{U \cap \partial M}) : (U, \varphi) \in \mathcal{A}\} \end{align*} (discarding empty domains) covers $\partial M$ because every boundary point of $M$ lies in some boundary chart by definition. Each chart in $\mathcal{A}'$ has image open in $\mathbb{R}^{n-1}$ by Step 3. **Smoothness of transitions.** Given two atlas charts $(U, \varphi), (V, \psi) \in \mathcal{A}$ on $M$, the transition $\psi \circ \varphi^{-1}$ is smooth — meaning it extends to a smooth map on an open neighbourhood in $\mathbb{R}^n$. Restricting this smooth $\mathbb{R}^n$-extension to the affine subspace $\{x_n = 0\}$ (and then projecting out the trailing $0$ via $\pi$ and re-inserting via $y \mapsto (y, 0)$ on the source side) yields a smooth map between open subsets of $\mathbb{R}^{n-1}$. So $\mathcal{A}'$ is a smooth atlas. **The key observation.** Every chart in $\mathcal{A}'$ has image open in $\mathbb{R}^{n-1}$ — **not** in the half-space $\mathbb{H}^{n-1}$. Equivalently, every chart can be regarded as a chart with image in $\operatorname{Int} \mathbb{H}^{n-1}$ (when we view $\partial M$ as a smooth $(n-1)$-manifold-with-boundary). **Applying Step 2 to $\partial M$.** Step 2 established that the property "$\varphi(p) \in \partial \mathbb{H}^n$" is independent of the chart and characterises boundary points. Apply the same reasoning to the manifold-with-boundary $\partial M$ equipped with atlas $\mathcal{A}'$. Since every chart $(U', \varphi') \in \mathcal{A}'$ satisfies $\varphi'(U') \subseteq \mathbb{R}^{n-1} = \operatorname{Int} \mathbb{H}^{n-1}$, **no** point of $\partial M$ is sent to $\partial \mathbb{H}^{n-1}$ by any chart. By the intrinsic characterisation, no point of $\partial M$ is a boundary point of $\partial M$. Therefore \begin{align*} \partial(\partial M) = \varnothing. \end{align*} **Why this works in one breath.** The "boundary" operator $\partial$ records the obstruction to a point having a full Euclidean neighbourhood (as opposed to a half-Euclidean one). The boundary $\partial M$ inherits charts that are *already* full Euclidean — the half-space disappears the moment we restrict to the hyperplane $x_n = 0$. So there is no second-level obstruction left to detect, and $\partial(\partial M)$ has nothing to put in it.[/guided]