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 $\mathcal{U} \subseteq \mathcal{T}_{\mathrm{ind}}$ have union $X$. If $X = \varnothing$ 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 $\mathcal{T}_{\mathrm{Sier}}$ 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 $\mathcal{U} \subseteq \mathcal{T}_{\mathrm{cof}}$ have union $X \ne \varnothing$, 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 $\mathcal{T}_{\mathrm{coc}}$ 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 $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ 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 $\infty = \{\, n \in \mathbb{N} : n \notin n \,\} = \mathbb{N} = \omega$ by [L2], since every natural satisfies $n \notin n$. Hence $\mathbb{N}^{*} = \omega \cup \{\omega\} = \omega^{+} = \omega + 1$, 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 $\mathbb{N}^{*}$ are the subsets of $\mathbb{N}$ together with the sets $\mathbb{N}^{*} \setminus F$ 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 $\mathbb{R}^{*} \to S^1$. For $t \in \mathbb{R}$ one computes $(2t)^2 + (t^2-1)^2 = (t^2+1)^2$, so $h(t) \in S^1$, and $h(t) \ne (0,1)$ since $t^2 - 1 = t^2+1$ is impossible. Conversely for $(x,y) \in S^1$ with $y \ne 1$ put $t := x/(1-y)$; then $t^2 = x^2/(1-y)^2 = (1-y^2)/(1-y)^2 = (1+y)/(1-y)$, so $t^2+1 = 2/(1-y)$ and $t^2-1 = 2y/(1-y)$, whence $h(t) = (x,y)$; and $t$ is the unique such real, being recovered from $h(t)$ by the same formula. With $h(\infty) = (0,1)$ this makes $h$ a bijection.

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

$h$ is therefore a continuous bijection from $\mathbb{R}^{*}$, 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 $|t| < \iota(n)$ 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 $\{t : |t-p| \le 1\}$ 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 $\varepsilon > 0$ with $(r-\varepsilon, r+\varepsilon) \cap \mathbb{Q} \subseteq K$, 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 $r < t < r + \varepsilon$. Every neighbourhood of $t$ contains an interval $(c,d)$ with $r < c < t < d < r+\varepsilon$, and [L7] puts a rational $q$ with $c < q < d$ in it; that $q$ lies in $(r-\varepsilon, r+\varepsilon) \cap \mathbb{Q} \subseteq K$. 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: $\mathcal{B}_\ell$ 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 $\varnothing$ otherwise, so it is a member of $\mathcal{B}_\ell$ or empty. By [L1] the family is a basis for exactly one topology, and a set is open in $\mathbb{R}_\ell$ 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 $\mathcal{B}_\ell$, 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 $\mathbb{R}_\ell$ is not compact.

For claim 3 let $\mathcal{A}$ be an open cover of $\mathbb{R}_\ell$ and let $\mathcal{D}$ be the family of members of $\mathcal{B}_\ell$ 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 $\mathbb{R}_\ell \times \mathbb{R}_\ell$ the antidiagonal $\Delta$ is discrete as a subspace: for $x \in \mathbb{R}$ the basic set $[x, x+1) \times [-x, -x+1)$ meets $\Delta$ 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 $\Delta$ is therefore open in the subspace.

$\Delta$ is closed in $\mathbb{R}_\ell \times \mathbb{R}_\ell$: let $(u,v)$ have $u + v \ne 0$. 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 $\Delta$. If $u+v < 0$, put $\delta := -(u+v)/2 > 0$; 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 $\Delta$. So the complement of $\Delta$ is open.

$\mathbb{R} \setminus C$ is at most countable. Fix a surjection $\mathbb{N} \to \mathbb{Q}$ ([L4]) and for $x \notin C$ 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 $x$.