I was thinking of a way to prove this and I realised that for my approach the lemma from the title would be useful, and it´s an interesting question on its own. Obviously it is true if the manifold is compact or $\mathbb{R}^n$, but I don´t see it in general. This is the precise question:

Let $M$ be a connected manifold and $X$ a compact inside it, then is the union of $X$ and all the relatively compact components of $M\setminus X$ compact?

P.S. I already asked this question in MSE a few days ago, but after asking a few people I thought it could be more appropiate for MO.

  • 1
    $\begingroup$ I doubt that this is true. There are simple examples which fail to be manifolds, e.g., $X= [0,1]\times \{0\} \cup \{0\}\times [0,\infty) \cup \bigcup_{n\in\mathbb N} \{1/n\}\times [0,n]$ and $K=[0,1]\times [0,1]$. Perhaps one can make such a comb a manifold in higher dimensions. $\endgroup$ Dec 5 '21 at 12:05

I believe your set is indeed compact. Its complement is the collection of points $x$ such that there exists a continuous curve $\gamma^x:\mathbb R_+\to M$ such that

  • $\gamma^x(0)=x$;
  • $\gamma^x$ leaves every compact set;
  • the image of $\gamma^x$ stays outside $X$.

According to this point of view, it is clear that this set is open, and decreasing in $X$. It will suffice to show that your set is compact for some $X'\supset X$ large enough.

Let $X'\supset X$ be a compact submanifold with boundary in $M$ of maximal dimension (for instance, embed $M$ as a closed set of $\mathbb R^n$, and consider the intersection with a large close ball; by Sard's lemma it is a manifold with boundary most of the time). I claim that $M\setminus X'$ only has finitely many components, which will conclude. Every point of $\partial X'$ admits a neighbourhood that intersects exactly one connected component of $M\setminus X'$ (because locally $X'$ is just a half-space). By compactness, we can find an open neighbourhood of $\partial X'$ that intersects only finitely many components, and by connectedness of $M$ every such component intersects every neighbourhood of $\partial X'$, so those are finitely many as advertised.

  • 1
    $\begingroup$ I am using Sard's theorem in the following way: 1) since the function $x\mapsto \|x\|^2$ is smooth, its regular values are dense 2) if $c$ is a regular value of a smooth map $f:M\to\mathbb R$, then $X'=f^{-1}(-\infty,c]$ is a submanifold with boundary, and $\partial X'=f^{-1}\{c\}$. $\endgroup$
    – Pierre PC
    Dec 5 '21 at 16:08
  • $\begingroup$ Thanks! I noticed after writing the comment $\endgroup$ Dec 5 '21 at 16:11

Actually, there is an elementary proof of this. I will imitate the one given in O.Forster Lectures on Riemann Surfaces. We assume $M$ is connected. Let $Y$ be equal to the union of $X$ with all the relatively compact components of $M \setminus X$

Let $U$ be a relatively compact, open subset containing $X$ and let $C_j$, $j \in J$ be the connected components of $M \smallsetminus X$. Let $bU$ be the boundary of $U$, which is compact and disjoint from $X$.

  • Claim 1. Every $C_j$ meets $U$: If $C_j \subset M \setminus U$, its closure in $M$ is also contained in $M \setminus U$, but $C_j$ is a connected component of $M \setminus U$ so $C_j = \overline{C_j}$ which conttradicts connectedness of $M$.
  • Claim 2. Only finitely many $C_j$ intersects $bU$: This is because $bU$ is compact and the $C_j$ are open and disjoint, anc cover $bU$.
  • Claim 3. $Y$ is closed: Let $J_0$ consist of the indices corresponding to relatively compact components. Then $M \setminus Y = \bigcup_{j \not \in J_0} C_j$, which is open.

By Claim 2, we can find $j_1, \ldots , j_k \in J_0$ be such that $C_{j_i}$ intersects $bU$. Then , by Claim 1 again, any other $C_j$ is contained in $U$. Therefore, $$ Y \subset U \cup C_{j_1} \cup \ldots \cup C_{j_k}$$ RHS is relatively compact by the choice of $U$ and the $j_i$, and LHS is closed by Claim 3, so LHS is compact too.

  • $\begingroup$ This is a wonderful proof! What do we need about $M$, it should be Hausdorff, locally compact, locally connected? This would rule out Jochen Wengenroth's example above. $\endgroup$
    – Pierre PC
    Dec 11 '21 at 9:48
  • $\begingroup$ @PierrePC I think you also need $M$ to have only finitely many connected components, if not to be connected itself. Example: $M=\mathbb N$ with the discrete topology is LCH and locally connected, but if $X=\{1\}$ then $Y = \mathbb N$ is not compact. $\endgroup$ Dec 11 '21 at 18:00
  • $\begingroup$ Ah yes, I see. I would say the remaining components are rather uninteresting though. I would probably consider only the inverse image of the interesting component in $M/K$ or so. $\endgroup$
    – Pierre PC
    Dec 13 '21 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.