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Based on this question I'd like to know: Are there compact (sub)manifolds without boundary in $\mathbb{R}^n$ ? Because, as that question shows, the topology of the manifolds has to be the trace topology; thus compact subspace (in particular, compact manifolds) are characterized by the Heine-Borel theorem: They are precisely those sets in $\mathbb{R}^n$ that are closed and bounded.

But, as far as I know (and I'm just starting to read about manifolds and haven't got a good grasp on the formal definitions yet), manifolds without boundary aren't closed, so they can't be compact ?

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    $\begingroup$ The boundary of a compact $n$-manifold, if not empty, is a compact $n-1$-manifold. But boundaries have no boundary. $\endgroup$ – ajotatxe May 12 '15 at 15:11
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    $\begingroup$ Think of $S^{n-1} \subseteq \mathbb{R}^n$ defined as the set of vectors of length $1$. This is clearly a closed and bounded set of $\mathbb{R}^n$, and yet $S^{n-1}$ has no boundary. $\endgroup$ – Dorebell May 12 '15 at 15:11
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    $\begingroup$ Every compact submanifold of $S^{n-1}$ is in particular a closed subset of $S^{n-1}$ and thus (since $S^{n-1}$ is closed in $\mathbb{R}^n$) a closed subset of $\mathbb{R}^n$. $\endgroup$ – Dorebell May 12 '15 at 15:54
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$\newcommand{\Reals}{\mathbf{R}}$Comparing this question with your linked question, the central issue seems to be the term "closedness", which perhaps feels bothersome because manifolds are unions of open sets.

If that's really the question, the resolution comes down to "relative topology", how "open" and "closed" are defined for subsets of $\Reals^{n}$.

If $X \subset \Reals^{n}$ is an arbitrary non-empty set, we say $A \subset X$ is (relatively) open in $X$ if there exists an open set $U \subset \Reals^{n}$ such that $A = X \cap U$. Relatively closed sets are defined similarly.

In particular, every set $X$ is both (relatively) open and closed as a subset of itself: The set $U = \Reals^{n}$ is both open and closed, and $X = X \cap \Reals^{n}$.

Now let's take an example of a compact manifold in $\Reals^{n}$, such as the $(n - 1)$-sphere $$ S^{n-1} = \{x \text{ in } \Reals^{n} : \|x \| = 1\}. $$ As the zero set of a continuous function $f:\Reals^{n} \to \Reals$, the sphere $S^{n-1}$ is closed in $\Reals^{n}$. Certainly, $S^{n-1}$ is not open in $\Reals^{n}$; in fact, $S^{n-1}$ has empty interior in $\Reals^{n}$.

As noted above, however, $S^{n-1}$ is both open and closed as a subset of itself. There's no contradiction because "open" and "closed" are relative concepts. (Remarkably and not completely obviously, "compactness" of $X$ is an intrinsic concept, not depending on how $X$ is viewed as a subset of a larger universe.)

Going back to the sphere, the hemispheres $$ H_{i}^{+} = \{x \text{ in } S^{n-1} : x_{i} > 0\},\qquad H_{i}^{-} = \{x \text{ in } S^{n-1} : x_{i} < 0\} $$ are (relatively!) open subsets of $S^{n-1}$ (why?) constituting a covering by coordinate charts. It follows that $S^{n-1}$ is a compact manifold in $\Reals^{n}$.

(Incidentally, this is an "inefficient" covering of the sphere; stereographic projection allows the sphere to be covered with two coordinate charts, each covering the complement of one point.)

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  • $\begingroup$ Compactness comes from Heine-Borel: The sphere is a closed, bounded subset of $\Reals^{n}$. A covering by chart neighborhoods shows the sphere is a manifold (and since we know $S^{n-1}$ is compact, the covering shows $S^{n-1}$ is a compact manifold). :) $\endgroup$ – Andrew D. Hwang May 13 '15 at 10:19
  • $\begingroup$ One last question, to make sure I understood: Does that mean that if we don't allow (sub)manifolds with boundary, then except $S^{n-1}$ itself (and perhaps $\emptyset$ which I shall exclude from consideration as a pathologicality) there aren't any compact submanifolds of $S^{n-1}$ ? $$$$Because every submanifold is obtained by intrsecting an $R^n$-open set with $S^{n-1}$, but a set obtained in such a way is not closed, thus not compact.$$$$If we do allow (sub)manifolds wit boundary, the intersection of closed sets, like $R\times [0,1]$ with $S^{n-1}$, would yield a compact submanifold, right ? $\endgroup$ – temo May 13 '15 at 10:53
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    $\begingroup$ If $M$ is a manifold, then every open subset of $M$ is a submanifold ("usually" non-compact). However, $M$ may also contain compact, boundaryless submanifolds of smaller dimension. The complement of a point in $S^{n-1}$ is diffeomorphic to $\Reals^{n-1}$, so every compact, boundaryless submanifold in $\Reals^{n-1}$ gives a compact, boundaryless submanifold in $S^{n-1}$. For instance, if $n \geq 2$, there are copies of $S^{n-2}$ inside $S^{n-1}$. $\endgroup$ – Andrew D. Hwang May 13 '15 at 11:05
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    $\begingroup$ Two more notes on terminology: 1. "Bounded" and "with boundary" have distinct meanings in this context. A subset $X$ of $\Reals^{n}$ is "bounded" if $X$ is contained in some ball, whether or not $X$ is a submanifold. "With boundary" unambiguously refers to a manifold-with-boundary, and usually suggests the boundary is non-empty. 2. If $X$ is a manifold, a "closed submanifold of $X$" is a submanifold that is closed in the topology of $X$ (logical); but to say "$X$ is a closed manifold" almost always means $X$ is compact and boundaryless (convenient, but vexing to the unwary). $\endgroup$ – Andrew D. Hwang May 13 '15 at 11:17
  • $\begingroup$ Ah, know now where my confusion came from (But first: Thank you for your detailed comments and fine explanations!). I wrote "$S^{n-1}$" but in my head I pictured $S^1$ embedded in $\mathbb{R}^2$. In this special case, I believe my observations from above should hold true, right ? $\endgroup$ – temo May 14 '15 at 9:27
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Well there are several examples. @Dorebell mentioned one example. Here are some other compact manifolds without boundary:

  • Torus
  • Double Torus
  • Klein bottle
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The Riemann sphere is a compact complex manifold without boundary.

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  • $\begingroup$ Certainly the set of unit vectors in $\mathbf{R}^{3}$ is a compact manifold (diffeomorphic to the Riemann sphere), but calling this surface the Riemann sphere is perhaps unnecessary (since the OP is a beginner) and arguably misleading (since no compact complex manifold of positive dimension embeds holomorphically in any complex Euclidean space, by the maximum principle). :) $\endgroup$ – Andrew D. Hwang May 12 '15 at 15:18
  • $\begingroup$ You make a good point. $\endgroup$ – Tim Raczkowski May 12 '15 at 15:42
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Yes; there are some. If $M$ is a compact, oriented $n$-manifold without boundary, then there must be some $n$-forms which do not arise from taking the exterior derivatives of $n-1$-forms.

If $M$ is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on $M$ which are not the divergence of any vector field on $M$.

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    $\begingroup$ This looks very very interesting but I don't easily see how it answers the question... more like a teaser :) Is this hinting to use the Morse theory? For people like me would you be so kind as to support this with a couple of references? $\endgroup$ – rych Sep 6 '15 at 5:31
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A compact manifold without a boundary is called a closed manifold so it is certainly an important class of manifolds.

Example: A $n-1$sphere ${\cal S}^{n-1}$ is a closed manifold, but the unit ball enclosed $\{\mathbf{x}\in \mathbb{R}^{n}\|\mathbf{x}\leq 1\|\}$ is a campact manifold with its boundary being ${\cal S}^{n-1}$.

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