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In the definition of a manifold with boundary, we often work with the closed upper half-space $H^{n}=\left\{ x\in\mathbb{R}^{n}\,|\, x^{n}\geq0\right\}$ , and endow $H^{n}$ with the subspace topology inherited from $\mathbb{R}^{n}$ , which we can call $\tau$ . We then say that the boundary of $H^{n}$ is given by $\partial H^{n}=H^{n}=\left\{ x\in\mathbb{R}^{n}\,|\, x^{n}=0\right\}$ . However, if we're working with the subspace topology, does this not imply that $interior\left(H^{n}\right)=H^{n}$ , since $H^{n}$ would itself be open in this topology? It then follows that $\partial H^{n}=\emptyset$ , which doesn't make sense in our context. Are we instead using the $\partial H^{n}$ from the original Euclidean topology and carrying that language over by some convention? Thanks a lot.

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First of all, there is no need to involve topology in the definition of $\partial H^n$, you can just define it as a set ($\{x^n=0\}$). If you want to define it using topological terms (the notation and the intended use indicate that this is desired) it is in fact the boundary of $H^n$ with respect to the topology of $\mathbb{R}^n$.

The boundary of a manifold is, in any case, by definition, the image of that set in coordinate charts. Any properties of this set are to be derived from that definition.

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  • $\begingroup$ Thomas, thanks a lot, it is much clearer now. $\endgroup$ – Mark Jan 9 '15 at 12:21

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