When it is known that a set $A$ in topological space $X$ can be exhausted by compact sets, that is there exists increasing sequence of compact sets covering $A$?
Comments:
I guess this should involve both conditions on topology of $X$ and the set $A$ itself.
I remember my lecturer in analysis using such exhaustion for open set in Euclidean space but I don't even know proof for that case.
I will appreciate both answers and references to sources where I can find it.
On
This question and its answer is actually a lot more subtle than acknowledged in the comments. There are at least two issues:
One can find at least two distinct definitions of “exhaustion by compact sets” in usage, and they are not equivalent.
At least one of the notions is a relative/extrinsic property (depending on an “ambient topology”), and not an intrinsic/topological property (in particular not necessarily preserved by homeomorphism).
First I will explain the two definitions and try to emphasize where they differ.
Next, I will point out how this difference is relevant to the distinction between relative (extrinsic) and topological (intrinsic) properties, giving an example.
Finally, I will relate all three definitions to other “compactness” concepts (sigma-compactness, hemicompactness, paracompactness, and local compactness).
Definitions:
Definition 1 (“Strong” Exhaustion by Compact Sets)
Given a set $A$ (within a topological space $X$), a sequence $\{K_n\}_{n=1}^{\infty}$ of compact sets exists such that $K_n \subseteq \operatorname{Int}_X K_{n+1}$ for all $n \ge 1$, and $\bigcup\limits_{n=1}^{\infty} K_n = A$.
Note that this implies that the (relative) interiors of the compact sets $K_n$ form an open cover of $A$.
Definition 2 (“Weak” Exhaustion by Compact Sets)
Given a set $A$ (within a topological space $X$), a sequence $\{K_n\}_{n=1}^{\infty}$ of compact sets exists such that $K_1 \subseteq K_2 \dots \subseteq K_n \subseteq \dots$ and $\bigcup\limits_{n=1}^{\infty} K_n = A$.
Cf. e.g. the usage here for definition 2, which, while seemingly less common than definition 1, appears to be what the comments refer to. (Because only definition 2 is equivalent to $\sigma$-compactness, see below.)
Clearly definition 1 implies definition 2. However, the converse is false and these two definitions are not equivalent (again see below).
Extrinsic vs. Intrinsic Properties:
Properties like connectedness or compactness are “intrinsic” or “topological” properties. They are preserved under homeomorphism, and remain unchanged regardless of what “ambient” or “external” topological space the space with the property in question might be embedded in. In particular, we can always assume without loss of generality that the space itself is the “ambient” or “external” topological space.
Related:
In contrast, properties like “open” or “closed” depend on the “ambient” or “external” topological space. For example, given an arbitrary subset $Y$ of a topological space $X$, it may be neither open nor closed. However, considered as a subset (of itself) in the subspace topology inherited from $X$, $Y$ will always be both closed and open. Therefore, there is never any “universal” or “intrinsic” sense in which a space is “open” or “closed”. Whenever we say a space has such a property, we must implicitly or explicitly be referring to some choice of “ambient” or “external” topology.
Cf. this answer to another question.
Relevance of Distinction to the notion of "interior":
Note that by definition, the interior $\operatorname{Int} Y$ of any set $Y$ is the largest open set contained within $Y$. Thus “interior” is to some extent a “weasel word”, because there is no “universal” or “intrinsic” sense in which a set is oepn, and thus no "universal" or "intrinsic" sense in which a set is the "largest open set" contained by, i.e. the interior of, $Y$. Instead, the definition of “interior of $Y$” depends inherently on the context of the “external” or “ambient” topological space containing $Y$.
Related questions:
For example, consider the integers $\mathbb{Z}$. Considered as a subset of R with the Euclidean topology, any point of $\mathbb{Z}$ is closed but not open, and thus being a singleton set the only open set it contains is the empty set. In other words, the interior of any point of $\mathbb{Z}$ is empty, when $\mathbb{Z}$ is considered within the “ambient” or “external” topology defined by $\mathbb{R}$.
In contrast, when $\mathbb{Z}$ is considered with the subspace topology it inherits from $\mathbb{R}$, which is the discrete topology, i.e. when $\mathbb{Z}$ is considered within the “ambient” or “external” topology defined by itself, every point of $\mathbb{Z}$ is both closed and open. Therefore, in this context, the interior of every point is the point itself.
Thus the definition of “interior” of a point of $\mathbb{Z}$ has no “universal” meaning, and is not preserved by the homeomorphic embedding of $\mathbb{Z}$ as a subset of $\mathbb{R}$ with the Euclidean topology.
Relevance of Distinction to "Exhaustion by Compact Sets"
Definition 2 of “exhaustion by compact sets”, the “weak” definition, is defined exclusively in terms of compact sets (and their union), so it is an intrinsic or “topological” property preserved by homeomorphisms. Thus, when considering “weak exhaustions by compact sets”, we can always assume without loss of generality that the “ambient” topological space is the space itself.
In contrast, definition of 1 of “exhaustion by compact sets”, the “strong” definition, refers to the “interiors” of the compact sets. Therefore whether a “strong exhaustion by compact sets” exists does depend on the “ambient” topological space – it is a “relative” or “extrinsic” property. (Note that this happens to be reflected in the way this question is asked.) When considering “strong exhaustions by compact sets”, we can never ignore the choice of “ambient” or “external” topology.
The integers $\mathbb{Z}$ embedded within $\mathbb{R}$ are again an example of this. Considered as a subspace of $\mathbb{R}$, $\mathbb{Z}$ (and all of its subsets) have empty interior. Therefore $\mathbb{Z}$ possess no “strong” exhaustion by compact sets. In contrast, when considering $\mathbb{Z}$ with the discrete topology/subspace topology inherited from $\mathbb{R}$ (thus $\mathbb{Z}$ is now the "ambient space"), all sets are open, so the points themselves are their own interiors, and thus because all finite subsets of any topological space are compact, a “strong” exhaustion by compact sets exists.
In fact, we can (and probably should) define at least two different versions of “strong” exhaustion by compact sets.
Definition 1a (“Strong” Exhaustion by Compact Sets Relative to Ambient Space $X$) = Definition 1
Given a set $A$ (within a topological space $X$), a sequence $\{K_n\}_{n=1}^{\infty}$ of compact sets exists such that $K_n \subseteq \operatorname{Int}_X K_{n+1}$ for all $n \ge 1$, and $\bigcup\limits_{n=1}^{\infty} K_n = A$.
Definition 1b (Intrinsic “Strong” Exhaustion by Compact Sets)
Given a topological space $A$, a sequence $\{K_n\}_{n=1}^{\infty}$ of compact sets exists such that $K_n \subseteq \operatorname{Int}_X K_{n+1}$ for all $n \ge 1$, and $\bigcup\limits_{n=1}^{\infty} K_n = A$.
That being said, technically Definition 1b is a special case of Definition 1a when the “ambient space” is just the space itself. Nevertheless Definition 1b should be preserved under homeomorphisms, whereas Definition 1a may or may not be true for two homeomorphic “versions of the same space”, depending on the corresponding embeddings into ambient spaces. Again, $\mathbb{Z}$ serves as an example – it satisfies Definition 1b but not Definition 1a when embedded in $\mathbb{R}$ “the standard way”.
The OP appears to ask for Definition 1a, which then is probably more difficult to answer than Definition 1b. (Both are certainly more difficult to characterize than Definition 2.) I could also comment about how e.g. one could define an “extrinsic” or “relative” version of “local compactness” that would not be preserved under homeomorphism (using "extrinsic open neighborhoods" that are both open in the ambient topology and restricted to being subsets of the subspace), again using $\mathbb{Z}$ as an example, but that is probably off-topic. The point being however that “standard”, i.e. “intrinsic” local compactness (defined in terms of a space’s own open neighborhoods, rather than those of an ambient space) is an intrinsic/topological property. The same reasoning should apply to “strong exhaustion by compact sets” as well.
Related questions:
The following paper has more about the relationship between relative properties and corresponding intrinsic versions (analogous to that between Definitions 1a and 1b):
A.V. Arhangel'skii,
Relative topological properties and relative topological spaces,
Topology and its Applications,
Volume 70, Issues 2–3,
1996,
Pages 87-99,
ISSN 0166-8641,
https://doi.org/10.1016/0166-8641(95)00086-0.
Relationships with other notions of compactness:
Like the other answers I will restrict attention only to the "intrinsic" definitions, Definition 2 and Definition 1b, just because that is simpler to do. Again this is somewhat inadequate because the OP fairly clearly asks about Definition 1a = Defintion 1, the "strong and extrinsic" definition.
$\sigma$-compactness:
In any case, Definition 2 ("weak") is equivalent to being $\sigma$-compact, cf. the argument here. Basically, any increasing sequence whose union is the entire space is obviously a sequence whose union is the entire space, and for the converse, given an arbitrary sequence whose union is the entire space it can be modified into an increasing sequence by using the "partial unions" of the first $n$ terms (because finite unions of compact sets are compact).
In particular, any conditions that imply $\sigma$-compactness must therefore imply Definition 2.
paracompactness:
Proposition 0.3 of the same notes shows that any space that is Hausdorff and satisfies definition 1b is paracompact. If we believe that Definition 1b is equivalent to (locally compact Hausdorff + $\sigma$-compact), see below, then there is another way to see that being Hausdorff and satisfying definition 1b implies paracompactness. Namely, any $\sigma$-compact space is Lindelöf (cf. Wikipedia), any locally compact Hausdorff space is regular (cf. Wikipedia), and any space that is both Lindelöf and regular is paracompact (you guessed it - cf. again Wikipedia). (Alternatively, cf. the answer to this question. Another related question is here.)
This answer to a related question can be thought of as showing how, using the fact that (locally compact + $\sigma$-compact) implies Definition 1b, those two conditions in turn imply paracompactness. Likewise Proposition 1.2 here also shows how (locally compact + $\sigma$-compact) implies paracompactness using Definition 1b as an intermediate.
hemicompactness:
Definition 1b is strictly stronger than Definition 2, because Definition 1b implies that the space is hemicompact. This is fairly easy to see, because the hypothesis implies that the interiors of the compact sets in the increasing sequence form an open cover of the entire space. Thus, given any compact subset $K$, we can take a finite subcover of that open cover, say $\{U_{n_1}, \dots, U_{n_m}\}$, where recall that $U_n = \operatorname{Int} K_n$, so then for the largest integer $N$ in the indices $\{n_1, \dots, n_m\}$ of that finite cover, $\{U_N\}$ is also a finite cover, i.e. $K \subseteq U_N \subseteq K_N$.
@HennoBrandsma discusses the relationship between Definition 1b and hemicompactness here. For example, it also contains another description of the proof that Definition 1b implies hemicompactness.
Anyway, the fact that Definition 1b implies hemicompactness, and that there exist hemicompact spaces that are not $\sigma$-compact, for example $\mathbb{Q}$,
means that Definition 1b is strictly stronger than Definition 2 (which is equivalent to $\sigma$-compactness).
As a partial aside, note that any first countable hemicompact space is locally compact, cf. for example here, or this question:
Most spaces one is likely to encounter “in practice” are first countable. Therefore the result that first-countable hemicompact spaces are locally compact implies that almost all hemicompact spaces one is likely to encounter “in practice” satisfy definition 1b, i.e. are sigma-compact and locally compact.
That being said, hemicompact and definition 1b are not equivalent, with hemicompact being strictly weaker (i.e. more general).
The only counterexample I could find (remember, it must not be first-countable, so therefore it must be “weird”) appears to be the so-called “sequential fan”. Cf. Example D of the paper
Frank Siwiec
Countable Spaces Having Exactly One Nonisolated Point. I
Proceedings of the American Mathematical Society
Volume 57, Number 2, June 1976, pp. 345-350
and the section 3 of
Gary Gruenhage and Glenn Hughes
Completeness Properties in the Compact-Open Topology on Fans
Houston Journal of Mathematics
Volume 41, Number 1, January 2015, pp. 321-327
(locally compact + $\sigma$-compact):
In any case, it is a well known result that any locally compact and $\sigma$-compact space satisfies Definition 1b. Cf. for example Lemma 1.1 here, the last proposition here, as well as this question asking for (and receiving) a proof of this result:
In fact, definition 1b should be equivalent to (locally compact + $\sigma$-compact) because, to paraphrase the argument for this given by @HennoBrandsma here the fact that the increasing union of compact sets equals the entire space, and that their interiors are an increasing union of open sets that are an open cover, means that every point in the space must lie in at least one of those open sets and thus have a compact neighborhood, thus the space must be locally compact. As explained before, it is trivial that Definition 1b implies Definition 2 (which is equivalent to $\sigma$-compactness). Regarding the equivalence of Definition 1b and (locally compact + $\sigma$-compact), cf. also this answer and this comment on this website.
As somewhat of a sidenote, recall that in general Lindelöf is strictly weaker (more general) than $\sigma$-compact. However, locally compact spaces that are Lindelöf are also $\sigma$-compact. Thus (locally compact + $\sigma$-compact) is equivalent not only to Definition 1b, but also to (locally compact + Lindelöf).
(locally compact + 2nd countable):
Finally, a well-known sufficient condition for either Definition 1b and/or (locally compact + $\sigma$-compact) is (locally compact + 2nd countable). There is a whole page about this sufficient condition on nLab, as well as (at least) three related questions about it on this website:
This is true in particular if $A = X$ is a topological manifold (or if $A$ is a submanifold of a manifold $X$, but the embedding has nothing to do here). I will sketch the proof and you feel free to relax the hypotheses as far as you can. I recall seeing this in my differential geometry course to prove the existence of a partition of unity.
Lemma. Let $M$ be a topological manifold (this means second-countable Hausdorff and locally homeomorphic to $\mathbb R^n$ for some fixed $n$). Then there exists a countable base of coordinate neighborhoods whose closure is compact.
Proof. By second countability, $M$ admits a countable base. Pick one such base $\mathcal B$. Let $\mathcal B' \subset \mathcal B$ be the subset of those $B \in \mathcal B$ that are contained in a coordinate patch $(U,\varphi)$ of the topological manifold $M$ and have compact closure. Of course $\mathcal B'$ is a base ; it suffices to transport the information from $\mathbb R^n$ to the manifold $M$. Since $\mathcal B'$ is a subbase of $\mathcal B$, $\mathcal B'$ is countable.
Theorem. Every manifold $M$ has a compact exhaustion $$ \varnothing \neq W_1 \subseteq \overline{W_1} \subseteq W_2 \subseteq \cdots $$ where the $W_i$ are open, the $\overline{W_i}$ are compact and their union is $M$.
Proof. Pick a countable base of open subsets of $M$ with compact closure. Write this as $\mathcal B = \{B_1,B_2,\cdots,B_n,\cdots \}$ (i.e. put a total order on the base indexed by the naturals) and set $W_k = \bigcup_{i=1}^k B_i$. We will pick a sequence $1 \le k_1 < k_2 < \cdots $ such that $\overline{W_{k_i}} \subseteq W_{k_{i+1}}$. We pick $k_1 = 1$. Assume $k_1,\cdots,k_j$ have been picked, since $\bigcup_{i \ge 1} B_i = M \supseteq \overline{W_{k_j}}$, there exists $k_{j+1} > k_j$ such that $\overline{W_{k_j}} \subseteq W_{k_{j+1}}$. Since $\bigcup_{j \ge 1} W_{k_j} = M$, the same property holds of the closures $\overline{W_{k_j}}$.
Hope that helps,