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If $M$ is a connected smooth manifold, then it is easy to show that there is a sequence of connected compact smooth submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=1}^\infty(M_i)^\circ$.

I would guess it should also be true that if $M$ is a connected topological manifold then there is a sequence of locally tame connected compact submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=1}^\infty(M_i)^\circ$. How would one try to prove such a statement? The only proof I know of the statement in the smooth category is to start with any exhaustion by open sets with compact closure and then "smooth" their boundaries. However, modifying an open set in a topological manifold so that its boundary is a tamely embedded codimension 1 submanifold seems much more delicate (and perhaps there is even an obstruction to doing it!).

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Since topological manifolds of dimension $\le 3$ are smoothable, the question is about manifolds of dimension $\ge 4$. Kirby and Siebenmann proved for $n\ge 6$ that every topological $n$-manifold admits a handle decomposition; this was extended to $n=5$ by Freedman and Quinn (I think, it is Quinn's paper "Ends of maps, III"). This applies to noncompact manifolds as well. Using this handle decomposition you can easily construct the required exhaustion (just use finitely many handles). This settles the problem in all dimensions but 4.

Handle decomposition is known to fail in dimension 4, but there is an alternative argument: Take $N^5=M^4\times R$, construct an exhaustion of $N$ as above by compact submanifolds $S_i$. Now, Quinn proved in 1988 a topological transversality theorem in all dimensions ("Topological transversality holds in all dimensions"), which allows you to perturb each $S_i$ to $S_i'$ whose boundary is transversal to $M\times 0$. Then $S_i'\cap M$ will be the required exhaustion.

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    $\begingroup$ Ricardo: It is in Kirby-Siebenmann except they assumed $n\ge 6$. The trouble is that Kirby-Siebenmann book is essentially unreadable (for most of us). I understood it (a bit) only after Kirby's lectures in Davis few years ago. The advantage of Quinn's paper is that he at least clearly states the results... $\endgroup$
    – Misha
    Commented May 1, 2013 at 22:17
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    $\begingroup$ It seems that Quinn's theorem in "Topological transversality holds in all dimensions" allows for a proof of the result which does not directly invoke the existence of handle decompositions. First give a proper embedding $f:M\to\mathbb{R}^N$. Then use Quinn's result to deform $f$ to some embedding $f_i :M\to\mathbb{R}^n$ which is transverse to $\partial \overline{B_i(0)}$: according to Quinn, we may assume that $f$ and $f_i$ coincide outside a small neighbourhood of $\partial \overline{B_i(0)}$. Then $M$ is exhausted by the compact locally flat submanifolds $M_i=(f_i)^{-1}(\overline{B_i(0)})$. $\endgroup$
    – Ricardo Andrade
    Commented May 1, 2013 at 22:41
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    $\begingroup$ Misha, it wasn't me. $\endgroup$
    – Igor Belegradek
    Commented May 1, 2013 at 23:01
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    $\begingroup$ Oh, maybe it was me then. $\endgroup$
    – Misha
    Commented May 1, 2013 at 23:44
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    $\begingroup$ Here's an easier (in the sense of quoting fewer results) argument in dimension 4. There is nothing to prove for compact manifolds. In the same paper (Ends of Maps III: Dimensions 3 and 4, JDG 17 (1982)) in which he proved the existence of handle structures and gave transversality results, Quinn proved that non-compact 4-manifolds are smoothable. Hence your favorite method for smooth manifolds will work. In real terms, this isn't any easier, since this argument and the one Ricardo gives depend on essentially the same set of ideas. $\endgroup$
    – Danny Ruberman
    Commented May 2, 2013 at 17:50
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Whitney embedding theorem shows that any connected smooth manifold $M$, compact or not, admits a proper imbedding $\newcommand{\bR}{\mathbb{R}}$ into an Euclidean space $\bR^N$ where properness signifies that the intersection of the image of the embedding with any compact set is a compact set.

Assume that $M\subset \bR^N$ is properly embedded. For a point $q\in \bR^N$ define $f_q:M\to\bR$ by setting

$$ f_q(p)= |p-q|^2,\;\;\forall p\in M. $$

Since $M$ is properly embedded we deduce that sublevel sets $\lbrace f_q\leq c\rbrace\subset M$ are compact for any $t\in\bR$.

For generic $q\in\bR^N$ the function $f_q: M\to\bR$ is Morse. Fix such a $q$. Thus each sublevel set $\lbrace f_q\leq t\rbrace$ contains finitely many critical points. This implies that the set of critical values of $f$ is a discrete countable subset of $\bR$.

Choose an increasing and unbounded sequence $(r_n)_{n\geq 1}$ of regular values of $f_q$ and set

$$M_n:=\lbrace f_q\leq r_n\rbrace. $$

The collection $(M_n)_{n\geq 1}$ is an exhaustion of $M$ by compact manifolds with boundary.

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    $\begingroup$ This answer doesn't seem to address the question asked . . . $\endgroup$
    – John Pardon
    Commented May 1, 2013 at 21:50
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    $\begingroup$ Dear unknown: I think, the point of Liviu's answer was to give a clean and quick proof in the smooth case by a method different from the one you had in mind. Of course, I agree, he does not address the topological question. $\endgroup$
    – Misha
    Commented May 1, 2013 at 22:19
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    $\begingroup$ This is actually the idea behind handle body decomposition theorem. $\endgroup$
    – Bombyx mori
    Commented May 2, 2013 at 5:53
  • $\begingroup$ Sorry if this is stupid, but wouldn't any point $q$ work? The set of critical values might not be discrete in general, but it would still have measure 0, wouldn't it? Am I missing something? $\endgroup$
    – asymmetriad
    Commented Mar 11, 2023 at 9:01
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Doesn't this depend on the definition of "manifold". If the only condition is being locally Euclidean, then there are connected non second countable examples (e.g., the "long line") for which the answer to the question is clearly negative.

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    $\begingroup$ Yes, currently there seems to be a disagreement about 2nd countability axiom as a part of the definition of a manifold. Nevertheless, the 1st paragraph of the question indicates that the author regards 2nd countability as a requirement. Personally, I am troubled by the fact that we do not even have a classification of connected surfaces if the 2nd countability is dropped! (Under the 2nd countability condition, see mathoverflow.net/questions/4155 .) $\endgroup$
    – Misha
    Commented May 1, 2013 at 18:32

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