9
$\begingroup$

Everyone knows the Heine-Borel theorem characterizing compact subsets of Euclidean space. For any $n \in \mathbb N$ a set $A \subseteq \mathbb R^n$ is compact just in case it is closed and bounded (in the sense that it is contained in some large enough finite diameter ball).

There is a similar theorem for the irrationals, due to Rothberger. The irrationals are homeomorphic to the space $\mathbb N^\mathbb N$ of functions $f:\mathbb N \to \mathbb N$ topologized by the product topology of the discrete topology on $\mathbb N$. With this presentation a set $A \subseteq\mathbb N^\mathbb N$ is compact just in case it is closed and bounded where bounded means there is an $f:\mathbb N \to \mathbb N$ so that for all $g \in A$ we have $\forall n \, g(n) \leq f(n)$.

My question is whether there is a similar (or at least relatively neat) characterization of compact sets of rationals (with the subspace topology). Does some relatively simple (topological, geometric, combinatorial) property characterize when a closed $A \subseteq \mathbb Q$ is compact?

$\endgroup$
5
  • $\begingroup$ In RCA$_0$ every compact set of rationals is closed and bounded. $\endgroup$ Commented Feb 23, 2022 at 11:09
  • $\begingroup$ A subset of the rationals (or of any other metric space) is compact if and only if it is complete and totally bounded, but I suppose this is not the kind of characterization you are looking for? $\endgroup$ Commented Feb 23, 2022 at 11:15
  • 8
    $\begingroup$ Up to homeomorphism, every countable compact Hausdorff space is a countable successor ordinal. Conversely, it's easy to embed all countable limit ordinals in the rationals by transfinite induction, and successor ordinals will give you compact sets. $\endgroup$
    – Ville Salo
    Commented Feb 23, 2022 at 11:37
  • 2
    $\begingroup$ A subsete of $\mathbb Q$ is compact if and only if it is compact as a subset of $\mathbb R$ (compactness is an internal property). $\endgroup$ Commented Feb 23, 2022 at 13:29
  • $\begingroup$ @VilleSalo This was exactly the kind of answer I was looking for. Thanks. If you post it as an answer I would accept it. $\endgroup$ Commented Feb 24, 2022 at 8:32

3 Answers 3

16
$\begingroup$

It is easy to describe these up to homeomorphism:

  • Every countable compact Hausdorff space is homeomorphic to a countable successor ordinal, see Milliet - A remark on Cantor derivative.

  • Conversely, it's easy to embed all countable limit ordinals in the rationals by transfinite induction, and successor ordinals will give you compact sets.

$\endgroup$
2
  • $\begingroup$ I'm guessing this embedding does not need to be strictly monotone. Can anything interesting be said about cases in which it is or in which it's not? $\endgroup$ Commented Feb 26, 2022 at 22:36
  • 1
    $\begingroup$ @MichaelHardy. In my second answer, I explained that if we take the upper limit topology on $\mathbb{Q}$ (which is homeomorphic to the order topology on $\mathbb{Q}$), then the embedding of an ordinal into $\mathbb{Q}$ is always monotonic. $\endgroup$ Commented Feb 27, 2022 at 2:08
5
$\begingroup$

The characterization of the compact subspaces of $\mathbb{R}^{n}$ as closed and bounded sets can be extended in a natural way to the much more general context of uniform spaces and subspaces of complete uniform spaces; a subset of a complete uniform space is compact if and only if it is closed and totally bounded, and $\mathbb{Q}$ can be made into a complete uniform space.

Observe that we cannot give $\mathbb{Q}$ a complete metric that is compatible with the topology on $\mathbb{Q}$ since the Baire category theorem applies to all complete metric spaces, but it does not apply to $\mathbb{Q}$, so we really need to go all the way to uniform spaces in order to say that $\mathbb{Q}$ is complete.

A separated uniform space is a pair $(X,\mathcal{U})$ where $X$ is a set and $\mathcal{U}$ is a filter on $X\times X$ that satisfies the following properties:

  1. $R\in\mathcal{U}\Rightarrow R^{-1}\in\mathcal{U}$

  2. $\bigcap\mathcal{U}=\{(x,x)\mid x\in X\}$

  3. $\forall R\in\mathcal{U}\exists S\in\mathcal{U}:S\circ S\subseteq R$

We shall assume all uniform spaces are separated. For example, every metric space $(X,d)$ is a uniform space where the uniformity $\mathcal{U}$ is the filter generated by the sets $R_{\delta}=\{(x,y)\in X^{2}\mid d(x,y)<\delta\}$ (in fact,every uniform space can be obtained in a similar manner if we use a collection of possibly uncountably many pseudometrics instead of metrics)

A uniform space $(X,\mathcal{U})$ is said to be totally bounded if for each $R\in\mathcal{U}$, there are $x_{1},\dots,x_{n}\in X$ with $X=R[x_{1}]\cup\dots\cup R[x_{n}]$.

A uniform space $(X,\mathcal{U})$ is said to be complete if whenever $e:(X,\mathcal{U})\rightarrow(Y,\mathcal{V})$ is an embedding of uniform spaces with $e[X]$ dense in $Y$, the mapping $e$ is already a uniform homeomorphism (alternatively, a uniform space is complete if and only if every Cauchy filter converges). A uniform space $(X,\mathcal{U})$ is said to be non-Archimedean if $\mathcal{U}$ is generated by equivalence relations.

Theorem: A uniform space $(X,\mathcal{U})$ is totally bounded if and only if its completion is compact.

Theorem: A uniform space $(X,\mathcal{U})$ is compact in the induced topology if and only if it is complete and totally bounded.

Corollary: A subset $A$ of a complete uniform space $(X,\mathcal{U})$ is compact if and only if it is closed and totally bounded.

A topological space is said to be realcompact if it can be embedded as a closed subspace of $\mathbb{R}^{I}$ for some $I$. A topological space is said to be $\mathbb{N}$-compact if it can be embedded as a closed subspace of $\mathbb{N}^{I}$ for some $I$. A Hausdorff space is said to be ultraparacompact if every open cover has a refinement that is a partition of the space into open sets. A Hausdorff space is said to be zero-dimensional if it has a basis of clopen sets.

Theorem: Suppose that $X$ is a completely regular space. Then each of the following statements implies the next one:

  1. $X$ is second countable.

  2. $X$ is Lindelof and has cardinality below the first uncountable measurable cardinal.

  3. $X$ is paracompact and has cardinality below the first uncountable measurable cardinal.

  4. $X$ is realcompact.

  5. $X$ can be endowed with a compatible complete uniformity.

Theorem: Suppose that $X$ is a zero-dimensional space. Then each of the following statements implies the next one.

  1. $X$ is second countable.

  2. $X$ is Lindelof.

  3. $X$ is ultraparacompact and has cardinality below the first uncountable measurable cardinal.

  4. $X$ is $\mathbb{N}$-compact.

  5. $X$ can be endowed with a compatible complete non-Archimedean uniformity.

  6. $X$ can be written as an inverse limit of discrete spaces $(X_{d})_{d\in D}$ where all the transitional mappings and the mapping from $X$ to $X_{d}$ are surjective.

Therefore, the space $\mathbb{Q}$ can be endowed with a complete uniformity $\mathcal{U}$, so a subspace $A\subseteq\mathbb{Q}$ is compact if and only if it is closed in $A$ and totally bounded with respect to this uniformity.

Suppose that $X$ is an inverse limit of an inverse system $(X_{d})_{d\in D}$ of discrete spaces. Then we have a characterization of the compact subspaces of $(X_{d})_{d\in D}$ that looks exactly like the characterization of the compact subsets of the irrational numbers. Here, we set $D$ to be downwards directed. If $d\leq e$, then let $\phi_{d,e}:X_{d}\rightarrow X_{e}$ be the transitional mapping, and let $\phi_{d}:X\rightarrow X_{d}$ is the canonical projection mapping. Suppose that each $\phi_{d,e},\phi_{d}$ is surjective. Then a closed subspace $C$ of $X$ is compact if and only if $\phi_{d}[C]$ is finite for each $d\in D$.

Let $\mathcal{K}$ denote the collection of all compact subsets of $X$, and let $\mathcal{J}$ denote the collection of all systems $(C_{d})_{d\in D}$ where $C_{d}$ is a finite subset of $X_{d}$ for $d\in D$ and where $\phi_{d,e}[C_{d}]=C_{e}$ whenever $d\leq e$. Observe that if $C_{d}$ is a finite subset of $X_{d}$ for each $d\in D$, then $\bigcap_{d\in D}\phi^{-1}[C_{d}]$ is closed and totally bounded, so $\bigcap_{d\in D}\phi^{-1}[C_{d}]$ is compact. Define a mapping $L:\mathcal{K}\rightarrow\mathcal{J}$ by letting $L(C)=(\phi_{d}[C])_{d\in D}$ and define a mapping $M:\mathcal{J}\rightarrow\mathcal{K}$ by letting $M((C_{d})_{d\in D})=\bigcap_{d\in D}\phi^{-1}_{d}[C_{d}]$. Observe that if $d\leq e$, then $\phi^{-1}_{d}[C_{d}]\subseteq\phi^{-1}_{e}[C_{e}]$, so the intersection $\bigcap_{d\in D}\phi^{-1}_{d}[C_{d}]$ is downwards directed.

Theorem: The mappings $L,M$ are inverses.

Proof: Suppose that $C$ is a compact subspace of $X$. Then $L(C)=(\phi_{d}[C])_{d\in D}$, so $M(L(C))=\bigcap_{c\in C}\phi_{d}^{-1}[\phi_{d}[C]]$. Since $C\subseteq\phi_{d}^{-1}[\phi_{d}[C]]$. On the other hand, if $x\in X\setminus C$, then since $C$ is closed, there is some $d\in D$ where $\phi_{d}(x)\not\in\phi_{d}[C]$. In this case, $x\not\in\phi_{d}^{-1}[\phi_{d}[C]]$. Therefore, $M(L(C))=C$.

Now, suppose that $(C_{d})_{d\in D}\in\mathcal{J}$. Then $M((C_{d})_{d\in D})=\bigcap_{d\in D}\phi_{d}^{-1}[C_{d}]$. Therefore, if $L(M((C_{d})_{d\in D}))=(A_{d})_{d\in D}$, then $$A_{d_{0}}=\phi_{d_{0}}[\bigcap_{d\in D}\phi_{d}^{-1}[C_{d}]]\subseteq\phi_{d_{0}}[\phi_{d_{0}}^{-1}[C_{d_{0}}]]\subseteq C_{d_{0}}.$$

On the other hand, if $c\in C_{d_{0}},$ then there is some $(c_{d})_{d\in D}\in\varprojlim_{d\in D}C_{d}$ with $c_{d_{0}}=c$ (this can be proven using Tychonoff's theorem and the fact that each $C_{d}$ is finite). In this case, $(c_{d})_{d\in D}\in\bigcap_{d\in D}\phi_{d}^{-1}[C_{d}]$. Therefore, since $c=\phi_{d_{0}}((c_{d})_{d\in D})$, we have $c\in\phi_{d_{0}}[\bigcap_{d\in D}\phi_{d}^{-1}[C_{d}]]$ as well. Therefore, $C_{d_{0}}\subseteq A_{c_{0}}$ as well, so $L(M((C_{d})_{d\in D}))=(A_{d})_{d\in D}=(C_{d})_{d\in D}.$ Q.E.D.

$\endgroup$
3
  • 1
    $\begingroup$ What does the inverse limit presentation of $\mathbb{Q}$ look like, concretely? Some kind of rounding to lattices of the form $r\mathbb{Z}$? And it's some kind of subtle limit of uniform spaces? $\endgroup$
    – Ville Salo
    Commented Feb 24, 2022 at 5:45
  • $\begingroup$ Thanks for this answer @JosephVanName. I'm not familiar with uniform spaces so I'll have to look into it more. $\endgroup$ Commented Feb 24, 2022 at 13:21
  • $\begingroup$ One such inverse system is the collection of all partitions of $\mathbb{Q}$ into clopen sets. The downwards directed set in the inverse system must be uncountable because inverse limits with countable index sets are completely metrizable and $\mathbb{Q}$ is not. $\endgroup$ Commented Feb 24, 2022 at 13:55
5
$\begingroup$

If we consider the ordering on $\mathbb{Q}$, then we obtain characterizations of the compact subsets of $\mathbb{Q}$ which give more information about these compact subsets than the homeomorphism type in Ville Salo's answer.

A linearly ordered set $X$ is said to be scattered if $X$ does not contain any order isomorphic copy of the rational numbers.

Let $\mathcal{A}_{\alpha}$ be the class of all totally ordered sets of the form $\bigcup_{a\in A}X_{a}$ where

  1. $A$ is either well-ordered or dual-well ordered,

  2. if $a<b$ and $x\in X_{a},y\in X_{b}$, then $x<y$, and

  3. either $X_{a}\in\bigcup_{\beta<\alpha}\mathcal{A}_{\beta}$ or $\lvert X_{a}\rvert\leq 1$.

Let $\mathcal{A}_{\infty}=\bigcup_{\alpha}\mathcal{A}_{\alpha}$.

Theorem: A linearly ordered set $X$ is scattered if and only if $X\in\mathcal{A}_{\alpha}$ for some ordinal $\alpha$.

Proof outline: The direction $\leftarrow$ is proven using a standard transfinite induction. For $\rightarrow$, suppose that $X\notin\mathcal{A}_{\alpha}$ for all ordinals $\alpha$. Then one can show that there is some $x\in X$ where $\{y\in X\mid y<x\}\notin\mathcal{A}_{\infty}$ and $\{y\in X\mid y>x\}\notin\mathcal{A}_{\infty}$. Therefore, by recursion, one can construct a set $X_{\gamma}$ along with an element $x_{\gamma}$ for each binary string $\gamma$ where we set $X_{\epsilon}=X$ and where for each binary string $X_{\gamma}$, there is an element $x_{\gamma}$ where if $X_{\gamma 0}=\{y\in X_{\gamma}\mid y<x\}$ and $X_{\gamma 1}=\{y\in X_{\gamma}\mid y>x\}$, then $X_{\gamma 0}\notin\mathcal{A}_{\infty}$ and $X_{\gamma 1}\notin\mathcal{A}_{\infty}$. In this case, the set $\{x_{\gamma}\mid \gamma\}$ is order isomorphic to $\mathbb{Q}$.

Q.E.D.

Proposition: A linearly ordered set $X$ is compact in the order topology if and only if $X$ is complete as a lattice.

Suppose that $A$ is a subset of a linearly ordered set $X$. Then an element $a\in A$ is said to be a gap point if there exists some $x\in X$ where either

  1. $a<x<\{b\in A\mid b>a\}$ but where the set $\{b\in A\mid b>a\}$ has no minimal element or

  2. $\{b\in A\mid b<a\}<x<a$ but where the set $\{b\in A\mid b<a\}$ has no maximal element.

Proposition: Suppose that $A$ is a subset of a linearly ordered set $X$. Then every set that is open in the order topology on $A$ is also open in the subspace topology.

Proposition: Suppose that $A$ is a subset of a linearly ordered set $X$. Then the following are equivalent:

  1. the order topology on $A$ coincides with the subspace topology.

  2. $A$ has no gap points.

  3. if $R\subseteq A$ and $\bigvee^{A}R$ exists, then $\bigvee^{X}R$ exists and $\bigvee^{A}R=\bigvee^{X}R$, and if $R\subseteq A$ and $\bigwedge^{A}R$ exists, then $\bigwedge^{X}R$ exists and $\bigwedge^{A}R=\bigwedge^{X}R$.

Proposition: Suppose that $X$ is a set and $\mathcal{S}$, $\mathcal{T}$ are compact Hausdorff topologies on $X$ with $\mathcal{S}\subseteq\mathcal{T}$. Then $\mathcal{S}=\mathcal{T}$.

Proposition: Suppose that $A$ is a subset of a linearly ordered set $X$. Then the following are equivalent:

  1. $A$ is a complete sublattice of $X$ in the sense that if $R\subseteq A$, then $\bigvee^{X}R,\bigwedge^{X}R$ exist and $\bigvee^{X}R,\bigwedge^{X}R\in A$.

  2. $A$ is compact in the subspace topology.

  3. $A$ is compact in the order topology, and the subspace topology on $A$ is the same as the order topology on $A$.

Let $X$ be a linearly ordered set. For each ordinal $\alpha$, let $\mathcal{C}_{\alpha,X}$ be the collection of all subsets $A\subseteq X$ where either

  1. there exists an ordinal $\delta$ and a continuous increasing function $f:\delta+1\rightarrow X$ where $f[\delta+1]\subseteq A$ and if $\gamma<\delta$, then $A\cap[f(\gamma),f(\gamma+1)]\in\bigcap_{\beta<\alpha}\mathcal{C}_{\alpha,X}$ or $\lvert A\cap[f(\gamma),f(\gamma+1)]\rvert=1$ and where $A\subseteq[f(0),f(\delta)]$, or

  2. there exists an ordinal $\delta$ and a continuous decreasing function $f:\delta+1\rightarrow X$ where $f[\delta+1]\subseteq A$, and if $\gamma<\delta$, then $A\cap[f(\gamma+1),f(\gamma)]\in\bigcap_{\beta<\alpha}\mathcal{C}_{\alpha,X}$ or $\lvert A\cap[f(\gamma+1),f(\gamma)]\rvert=1$ and where $A\subseteq[f(\delta),f(0)]$.

Let $\mathcal{C}_{\infty,X}=\bigcup_{\alpha}\mathcal{C}_{\alpha,X}.$

Proposition: A subset $A$ of a linearly ordered set $X$ is compact in the subspace topology and scattered if and only if $A\in\mathcal{C}_{\alpha,X}$ for some ordinal $\alpha$.

Proof:

$\leftarrow$ Observe that if $A\in\mathcal{C}_{\alpha,X}$ for some $A$, then $A\in\mathcal{A}_{\alpha}$, so $A$ is scattered. Furthermore, by transfinite induction, each $A\in\mathcal{C}_{\alpha,X}$ is a complete sublattice of $X$, so $A\in\mathcal{C}_{\alpha,X}$ is compact in the subspace topology.

$\rightarrow$ Suppose that $A\notin \mathcal{C}_{\infty,X}$. Then I claim that either $A$ is not compact in the subspace topology or $A$ has an isomorphic copy of the linear order $\mathbb{Q}$. Suppose that $A$ is compact in the subspace topology.

Then let $P=\{a\in A\mid\{b\in A\mid b\leq a\}\in\mathcal{C}_{\infty,X}\}$. Then $P$ has a largest element which we shall call $p$.

Let $Q=\{a\in A\mid\{b\in A\mid b\geq a\}\in\mathcal{C}_{\infty,X}\}$. Then $Q$ has a least element which we shall call $q$. If $q\leq p$, then we can conclude that $A\in\mathcal{C}_{\infty,X}$ which is a contradiction. We therefore know that $p<q$ and that $[p,q]\cap A\notin\bigcup_{\alpha}\mathcal{C}_{\alpha,X}$. In particular, if $x\in (p,q)\cap A$, then $(-\infty,x]\cap A\notin \mathcal{C}_{\infty,X}$ and $[x,\infty)\cap A\notin \mathcal{C}_{\infty,X}$.

Let $A_{\epsilon}=A$. We construct a system of elements $(x_{\gamma})_{\gamma\in\{0,1\}^{*}}$ in $A$ along with a system of subsets $(A_{\gamma})_{\gamma\in\{0,1\}^{*}}$ of $A$ recursively. For each binary string $\gamma$, let $x_{\gamma}\in A_{\gamma}$ be an element such that $[x_{\gamma},\infty)\cap A_{\gamma}\notin\mathcal{C}_{\infty,X}$ and $(-\infty,x_{\gamma}]\notin\mathcal{C}_{\infty,X}$. Then let $A_{\gamma 0}=(-\infty,x_{\gamma}]$ and let $A_{\gamma 1}=[x_{\gamma},\infty)$. Then $(x_{\gamma})_{\gamma\in\{0,1\}^{*}}$ is a subset of $A$ order isomorphic to $\mathbb{Q}$. Therefore, $A$ is not scattered.

Q.E.D.

Corollary: Let $X$ be a totally ordered set of cardinality less than the continuum. Then a subset $A\subseteq X$ is compact in the subspace topology if and only if $A\in\mathcal{C}_{\infty,X}$.

The upper limit topology

There is a simpler characterization of the compact subsets of $\mathbb{Q}$ where the subsets are given the subspace topology and $\mathbb{Q}$ is given the upper limit topology. Observe that since the upper limit topology on $\mathbb{Q}$ is homeomorphic to the order topology, the case when $\mathbb{Q}$ has the upper limit topology is not any different than the case with the order topology.

Recall that the upper limit topology on a totally ordered set $X$ is the topology where if $x$ is the least element of $X$, then $x$ is isolated, and where the local basis around $x$ consists of the intervals of the form $(v,x]$ where $v<x$.

Theorem: Every countable first countable regular space without any isolated points is homeomorphic to $\mathbb{Q}$ with the order topology.

Corollary: The set $\mathbb{Q}$ with the upper limit topology is homeomorphic to $\mathbb{Q}$ with the order topology.

Proposition: Let $X$ be a totally ordered set. Then $X$ is compact in the upper limit topology if and only if $X$ is well-ordered and has a greatest element.

Proposition: Let $X$ be a totally ordered set with the upper limit topology. Let $A$ be a subset of $X$, and give $A$ the subspace topology. Then $A$ is compact if and only if there is some ordinal $\alpha$, and a continuous function $f:\alpha+1\rightarrow X$ where if $\beta<\gamma\leq\alpha$, then $f(\beta)<f(\gamma)$.

$\endgroup$
1
  • $\begingroup$ At a certain point, you say "let $\mathcal C_{\alpha, X}$ be the collection of all subsets $A \subseteq X$ where either …", and then list two conditions; but the conditions are joined by "and". Should it be "or"? $\endgroup$
    – LSpice
    Commented Feb 26, 2022 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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