Compactness: Examples and Counterexamples

Claim 1. If $X$ is finite it is compact by [L1], whatever its topology. If $X$ is compact in the discrete topology, the family of singletons $\{\{x\}:x\in X\}$ is an open cover by [L2], so finitely many singletons cover $X$ and $X$ is listable, hence finite by [L3].

Claim 2. Let $\mathcal{U}\subseteq {\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}}$ have union $X$. If $X=\mathrm{\varnothing }$ the empty subfamily covers it; otherwise some member is nonempty, hence equals $X$ by [L2], and that single member is a finite subcover.

Claim 4. If $X$ is finite it is compact by [L1]. If $X$ is compact in ${\mathcal{T}}_p$, then $\{\{p,x\}:x\in X\}$ is a family of open sets by [L2] with union $X$, so finitely many of its members cover $X$; their union is a listable set, so $X$ is finite by [L3].

Claim 6. $S$ is finite, hence compact by [L1]; the subspace $\{b\}$ is a one-point space and so compact by [L1], making $\{b\}$ a compact subset by [L4]; and $\{b\}$ is not closed, its complement $\{a\}$ not lying in ${\mathcal{T}}_{\mathrm{S}\mathrm{i}\mathrm{e}\mathrm{r}}$ by [L2]. By [L5] this forces $S$ not to be Hausdorff, which it is not: the only open set containing $a$ is $S$.

Claim 3. Let $\mathcal{U}\subseteq {\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$ have union $X\mathrm{\ne }\mathrm{\varnothing }$, the empty case being as in step 1.2. Some $U_0\in \mathcal{U}$ is nonempty, so $X\setminus U_0$ is finite by [L2], say $X\setminus U_0=\{y_0,\dots ,y_k\}$ or empty; in the second case $\{U_0\}$ covers $X$, and in the first each $y_j$ lies in some member of $\mathcal{U}$, and finitely many members named in this way together with $U_0$ cover $X$.

Claim 5. Put $a_n:=\iota (n)$ for $n\in \mathbb{N}$ and $A_n:=\{a_m:m>n\}$, an at most countable subset of $\mathbb{R}$ by [L6] and [L3], so that $U_n:=\mathbb{R}\setminus A_n$ lies in ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{c}}$ by [L2]. Every real lies in some $U_n$: a real that is no $a_m$ lies in $U_0$, and $a_m$ lies in $U_m$. So $\{U_n:n\in \mathbb{N}\}$ is an at most countable open cover of $\mathbb{R}$.

The sets $U_n$ increase with $n$, since the $A_n$ decrease, so the union of finitely many of them is a single $U_N$, and $a_{N+1}\in A_N$ lies outside it. Hence this at most countable open cover has no finite subcover and $(\mathbb{R},{\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{c}})$ is neither countably compact nor compact, which is claim 5.

By [L3] the set $[0,1]$ is closed in $\mathbb{R}$ and bounded, so [L1] makes it a compact subset of the metric space $\mathbb{R}$ and [L2] makes it a compact subset of the topological space $\mathbb{R}$; that is, the subspace $[0,1]$ is a compact space, which is claim 1.

By [L4] the set $C$ is closed in $\mathbb{R}$ and contained in $[0,1]$, so $C=C\cap [0,1]$ is the trace of a closed set and hence closed in the subspace $[0,1]$; by step 1.1 that subspace is compact, so [L5] makes $C$ a compact subset of it, and by [L2] the subspace $C$ is a compact space, which is claim 2.

Each factor of $Q$ is the compact space $[0,1]$ of step 1.1, so [L6] makes $Q$ compact, which is claim 3.

For $X=\mathbb{N}$ the added point is $\mathrm{\infty }=\{n\in \mathbb{N}:n\mathrm{\notin }n\}=\mathbb{N}=\omega$ by [L2], since every natural satisfies $n\mathrm{\notin }n$. Hence ${\mathbb{N}}^{\ast }=\omega \cup \{\omega \}={\omega }^{+}=\omega +1$, an equality of sets and not merely a bijection.

For $X=\mathbb{R}$ the circle $S^1\subseteq {\mathbb{R}}^2$ carries the subspace topology of ${\mathbb{R}}^2$, which is metrizable by [L5]; so $S^1$ is metrizable and hence Hausdorff by [L5]. Nothing below uses compactness of $S^1$.

By [L3] the compact subsets of the discrete $\mathbb{N}$ are exactly its finite subsets, and every subset is closed; so by [L1] the open sets of ${\mathbb{N}}^{\ast }$ are the subsets of $\mathbb{N}$ together with the sets ${\mathbb{N}}^{\ast }\setminus F$ with $F\subseteq \mathbb{N}$ finite.

By [L4] a subset $V$ of $\omega +1$ is open in the order topology exactly when each of its points lies in a set $[0,\beta ]$ or $(\alpha ,\beta ]$ inside $V$. A subset of $\omega$ is open, each of its naturals $n$ lying in $[0,0]=\{0\}$ or in $(n-1,n]=\{n\}$; and a set $V\ni \omega$ is open exactly when it contains $[0,\omega ]=\omega +1$ or some $(\alpha ,\omega ]$ with $\alpha \in \omega$, that is exactly when $(\omega +1)\setminus V\subseteq [0,\alpha ]$ for some natural $\alpha$, that is exactly when its complement is finite.

$h$ is a bijection ${\mathbb{R}}^{\ast }\to S^1$. For $t\in \mathbb{R}$ one computes $(2t{)}^2+(t^2-1{)}^2=(t^2+1{)}^2$, so $h(t)\in S^1$, and $h(t)\mathrm{\ne }(0,1)$ since $t^2-1=t^2+1$ is impossible. Conversely for $(x,y)\in S^1$ with $y\mathrm{\ne }1$ put $t:=x\mathrm{/}(1-y)$; then $t^2=x^2\mathrm{/}(1-y{)}^2=(1-y^2)\mathrm{/}(1-y{)}^2=(1+y)\mathrm{/}(1-y)$, so $t^2+1=2\mathrm{/}(1-y)$ and $t^2-1=2y\mathrm{/}(1-y)$, whence $h(t)=(x,y)$; and $t$ is the unique such real, being recovered from $h(t)$ by the same formula. With $h(\mathrm{\infty })=(0,1)$ this makes $h$ a bijection.

$h$ is continuous at every point of $\mathbb{R}$: its two components are $t\mapsto 2t\mathrm{/}(t^2+1)$ and $t\mapsto (t^2-1)\mathrm{/}(t^2+1)$, quotients of polynomials whose denominator never vanishes, hence continuous by [L6], so $h$ restricted to $\mathbb{R}$ is continuous into ${\mathbb{R}}^2$ by [L6] and hence into the subspace $S^1$, which contains its image.

Claim 1 follows: by steps 2.1 and 2.2 the two topologies on the set $\omega +1$ of step 1.1 are the same family of subsets, so the identity map is a bijection carrying open sets to open sets in both directions and is a homeomorphism.

$h$ is continuous at $\mathrm{\infty }$. Let $V$ be open in $S^1$ with $(0,1)\in V$; by [L5] there is a real $r>0$ with every point of $S^1$ at $d_{\mathrm{\infty }}$-distance less than $r$ from $(0,1)$ lying in $V$. By [L8] fix a natural $M\ge 1$ with $1\mathrm{/}M<r\mathrm{/}2$, and put $C:=\{s\in \mathbb{R}:\mathrm{\mid }s\mathrm{\mid }\le M\}$, which is a closed bounded interval of $\mathbb{R}$, hence a compact subset of $\mathbb{R}$ by [L9]. For $t\mathrm{\notin }C$ one has $\mathrm{\mid }t\mathrm{\mid }>M\ge 1$, so $\mathrm{\mid }2t\mathrm{/}(t^2+1)\mathrm{\mid }\le 2\mathrm{\mid }t\mathrm{\mid }\mathrm{/}{t}^2=2\mathrm{/}\mathrm{\mid }t\mathrm{\mid }<2\mathrm{/}M<r$ and $\mathrm{\mid }(t^2-1)\mathrm{/}(t^2+1)-1\mathrm{\mid }=2\mathrm{/}(t^2+1)\le 2\mathrm{/}{t}^2\le 2\mathrm{/}M<r$; hence $h(t)\in V$. So $W:={\mathbb{R}}^{\ast }\setminus C$ is open in ${\mathbb{R}}^{\ast }$ by [L1], contains $\mathrm{\infty }$, and satisfies $h[W]\subseteq V$.

$h$ is therefore a continuous bijection from ${\mathbb{R}}^{\ast }$, which is compact by [L1], to $S^1$, which is Hausdorff by step 1.2; so [L7] makes it a homeomorphism, which is claim 2. With claim 1 at step 3.2 both statements are proved.

Each $I_n$ is closed in $\mathbb{R}$, its complement being the union of the open sets $\{t:t<-\iota (n)\}$ and $\{t:t>\iota (n)\}$, and it is bounded; so $I_n$ is a compact subset of $\mathbb{R}$ by [L2]. By [L3] every real $t$ satisfies $\mathrm{\mid }t\mathrm{\mid }<\iota (n)$ for some $n$, so $\mathbb{R}={\bigcup }_{n\in \mathbb{N}}{I}_n$, an at most countable union of compact subsets: claim 1.

Each singleton $\{r\}$ with $r\in \mathbb{Q}$ is a compact subset of $\mathbb{Q}$, the subspace it carries being a one-point space; and $\mathbb{Q}$ is the union of the family of its singletons, which is at most countable by [L4]. So $\mathbb{Q}$ is $\sigma$-compact: claim 2.

For claim 4 in $\mathbb{R}$: given $p\in \mathbb{R}$ the set $\{t:\mathrm{\mid }t-p\mathrm{\mid }\le 1\}$ is closed and bounded, hence compact by [L2], and it contains the open $(p-1,p+1)\ni p$, so it is a compact neighbourhood of $p$ and $\mathbb{R}$ is locally compact.

For claim 3 assume countable choice and let $\mathcal{U}$ be an open cover of $\mathbb{R}$. For $n\in \mathbb{N}$ the set $T_n$ of finite subfamilies of $\mathcal{U}$ covering $I_n$ is nonempty, $I_n$ being compact by step 1.1 and the ambient reading being licensed by [L9], so [L5] supplies ${\mathcal{V}}_n\in T_n$ for every $n$; the union ${\bigcup }_{n\in \mathbb{N}}{\mathcal{V}}_n$ is an at most countable subfamily of $\mathcal{U}$ by [L10], being a countable union of finite sets, and covers $\mathbb{R}$ by step 1.1. The same argument with the singletons of step 1.2 in place of the $I_n$ shows $\mathbb{Q}$ is Lindelöf: claim 3.

For claim 4 in $\mathbb{Q}$, let $r\in \mathbb{Q}$ and suppose $K\subseteq \mathbb{Q}$ were a compact neighbourhood of $r$ in $\mathbb{Q}$; then some set open in $\mathbb{Q}$ lies between $r$ and $K$, so by [L1] there is a real $\epsilon >0$ with $(r-\epsilon ,r+\epsilon )\cap \mathbb{Q}\subseteq K$, and by [L8] the set $K$ is a compact subset of $\mathbb{R}$ as well, hence closed in $\mathbb{R}$ by [L2].

By [L7] there is an irrational $t$ with $r<t<r+\epsilon$. Every neighbourhood of $t$ contains an interval $(c,d)$ with $r<c<t<d<r+\epsilon$, and [L7] puts a rational $q$ with $c<q<d$ in it; that $q$ lies in $(r-\epsilon ,r+\epsilon )\cap \mathbb{Q}\subseteq K$. So every neighbourhood of $t$ meets $K$, and $K$ closed gives $t\in K\subseteq \mathbb{Q}$ by [L8], contradicting the irrationality of $t$. Hence no point of $\mathbb{Q}$ has a compact neighbourhood in $\mathbb{Q}$, which completes claim 4.

Claim 1: ${\mathcal{B}}_{\mathrm{\ell }}$ covers $\mathbb{R}$, since $x\in [x,x+1)$; and $[a,b)\cap [c,d)$ is $[\max \{a,c\},\min \{b,d\})$ when that is nonempty and $\mathrm{\varnothing }$ otherwise, so it is a member of ${\mathcal{B}}_{\mathrm{\ell }}$ or empty. By [L1] the family is a basis for exactly one topology, and a set is open in ${\mathbb{R}}_{\mathrm{\ell }}$ exactly when each of its points has a half-open interval around it inside it.

Claim 2: the family $\{[-\iota (n),\iota (n)):n\in \mathbb{N},n\ge 1\}$ consists of members of ${\mathcal{B}}_{\mathrm{\ell }}$, hence of open sets, and covers $\mathbb{R}$ by [L3]; the members increase with $n$, so a finite subfamily has union $[-\iota (N),\iota (N))$ for the largest index $N$ occurring, which omits $\iota (N)$. So ${\mathbb{R}}_{\mathrm{\ell }}$ is not compact.

For claim 3 let $\mathcal{A}$ be an open cover of ${\mathbb{R}}_{\mathrm{\ell }}$ and let $\mathcal{D}$ be the family of members of ${\mathcal{B}}_{\mathrm{\ell }}$ contained in some member of $\mathcal{A}$; by step 1.1 the family $\mathcal{D}$ covers $\mathbb{R}$. Put $C:=\bigcup \{(a,b):[a,b)\in \mathcal{D}\}$.

In ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$ the antidiagonal $\mathrm{\Delta }$ is discrete as a subspace: for $x\in \mathbb{R}$ the basic set $[x,x+1)\times [-x,-x+1)$ meets $\mathrm{\Delta }$ only in $(x,-x)$, since a point $(y,-y)$ in it satisfies $x\le y$ and $-x\le -y$, that is $y\le x$. By [L7] each singleton of $\mathrm{\Delta }$ is therefore open in the subspace.

$\mathrm{\Delta }$ is closed in ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$: let $(u,v)$ have $u+v\mathrm{\ne }0$. If $u+v>0$, every point $(y_1,y_2)$ of $[u,u+1)\times [v,v+1)$ has $y_1+y_2\ge u+v>0$, so the box misses $\mathrm{\Delta }$. If $u+v<0$, put $\delta :=-(u+v)\mathrm{/}2>0$; every point of $[u,u+\delta )\times [v,v+\delta )$ has $y_1+y_2$ at least $u+v$ and less than $u+v+2\delta =0$, so again the box misses $\mathrm{\Delta }$. So the complement of $\mathrm{\Delta }$ is open.

$\mathbb{R}\setminus C$ is at most countable. Fix a surjection $\mathbb{N}\to \mathbb{Q}$ ([L4]) and for $x\mathrm{\notin }C$ let $r(x)$ be the rational of least index with $x<r(x)$ and $[x,r(x))\in \mathcal{D}$; such rationals exist, since $\mathcal{D}$ covers gives $[a,b)\in \mathcal{D}$ with $a\le x<b$, a rational $q$ with $x<q<b$ by [L3] then has $[x,q)\subseteq [a,b)$ and so $[x,q)\in \mathcal{D}$. Nothing is selected, the least index being determined by $x$. The map $r$ is injective on $\mathbb{R}\setminus C$: if $x<y$ lay outside $C$ with $r(x)=r(y)=q$, then $[x,q)\in \mathcal{D}$ gives $(x,q)\subseteq C$ and $x<y<q$ puts $y$ in $C$. So $\mathbb{R}\setminus C$ injects into $\mathbb{Q}$, hence is equinumerous with a subset of $\mathbb{Q}$ and at most countable by [L4].

$C$ is covered by the at most countable family ${\mathcal{D}}_{\mathbb{Q}}:=\{[p,q)\in \mathcal{D}:p,q\in \mathbb{Q}\}$, at most countable because $[p,q)\mapsto (p,q)$ injects it into $\mathbb{Q}\times \mathbb{Q}$, which is at most countable by [L4], as is therefore the image subset: given $x\in C$ there is $[a,b)\in \mathcal{D}$ with $a<x<b$, and [L3] gives rationals $p,q$ with $a<p<x<q<b$, whence $[p,q)\subseteq [a,b)$ lies in $\mathcal{D}$ and contains $x$.

So ${\mathcal{D}}_0:={\mathcal{D}}_{\mathbb{Q}}\cup \{[x,r(x)):x\in \mathbb{R}\setminus C\}$ is an at most countable subfamily of $\mathcal{D}$ by [L4] and covers $\mathbb{R}$ by steps 3.1 and 3.2. Every member of $\mathcal{D}$ lies inside some member of $\mathcal{A}$, so [L6] applied to an indexing of ${\mathcal{D}}_0$ by $\mathbb{N}$ supplies one member of $\mathcal{A}$ for each member of ${\mathcal{D}}_0$, and those form an at most countable subcover of $\mathcal{A}$. Hence ${\mathbb{R}}_{\mathrm{\ell }}$ is Lindelöf: claim 3.

$\mathrm{\Delta }$ is uncountable, being in bijection with $\mathbb{R}$ under $x\mapsto (x,-x)$ and $\mathbb{R}$ being uncountable by [L4]. Were ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$ Lindelöf, its closed subspace $\mathrm{\Delta }$ would be too: given a cover of $\mathrm{\Delta }$ by traces of ambient open sets, adjoining the complement of $\mathrm{\Delta }$ gives an ambient open cover, an at most countable subcover of it traces back to an at most countable subcover of $\mathrm{\Delta }$. But $\mathrm{\Delta }$ is discrete by step 2.3, so its singletons form an open cover admitting only itself as a subcover, and that family is uncountable. So ${\mathbb{R}}_{\mathrm{\ell }}\times {\mathbb{R}}_{\mathrm{\ell }}$ is not Lindelöf: claim 4.

The nonempty open subsets of $X$ are exactly the sets $U\times D$ with $U\subseteq \mathbb{N}$ nonempty: by [L1] and [L2] every basic open set is $U\times \mathrm{\varnothing }=\mathrm{\varnothing }$ or $U\times D$, and a union of sets of the second form is again of that form.

Every nonempty $A\subseteq X$ has a limit point in $X$. Take $(n,c)\in A$ and let $p:=(n,c^{\mathrm{\prime }})$ be the point with the same first coordinate and $c^{\mathrm{\prime }}\mathrm{\ne }c$, which exists since $D$ has two elements. Every neighbourhood of $p$ contains an open set containing $p$, hence by step 1.1 a set $U\times D$ with $n\in U$, and that set contains $(n,c)$, which lies in $A$ and differs from $p$. So $p$ is a limit point of $A$, and in particular every infinite subset of $X$ has one: $X$ is limit point compact.