The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies

Throughout, a topology is as in Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, and finite, at most countable and uncountable are as in Finite, countably infinite, countable, uncountable, so that "countable" always means "at most countable" and every finite set is countable. Let $X$ be a set. The six families below are topologies on $X$; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list.

1. Discrete topology. ${\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}}:=\mathcal{P}(X)$: every subset is open, hence every subset is closed, hence every subset is clopen.
2. Indiscrete topology. ${\mathcal{T}}_{\mathrm{i}\mathrm{n}\mathrm{d}}:=\{\mathrm{\varnothing },X\}$. Its closed sets are again $\mathrm{\varnothing }$ and $X$.
3. Cofinite topology. ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}:=\{\mathrm{\varnothing }\}\cup \{U\subseteq X:X\setminus U\text{ is finite}\}$. Its closed sets are $X$ together with the finite subsets of $X$.
4. Cocountable topology. ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{c}}:=\{\mathrm{\varnothing }\}\cup \{U\subseteq X:X\setminus U\text{ is at most countable}\}$. Its closed sets are $X$ together with the at most countable subsets of $X$.
5. Particular-point topology. Fix $p\in X$ and put ${\mathcal{T}}_p:=\{\mathrm{\varnothing }\}\cup \{U\subseteq X:p\in U\}$: the open sets are $\mathrm{\varnothing }$ and the sets containing $p$. Its closed sets are $X$ together with the sets not containing $p$.
6. Sierpinski topology. On a two-point set $S=\{a,b\}$ with $a\mathrm{\ne }b$, ${\mathcal{T}}_{\mathrm{S}\mathrm{i}\mathrm{e}\mathrm{r}}:=\{\mathrm{\varnothing },\{b\},S\}$. The pair $(S,{\mathcal{T}}_{\mathrm{S}\mathrm{i}\mathrm{e}\mathrm{r}})$ is Sierpinski space; $b$ is its open point and $a$ its closed point. This is exactly the particular-point topology of item 5 on a two-point set with particular point $b$, listed separately because it is quoted so often.

Two elementary facts about finite sets are used below, and both are proved here.

(i) A subset of a finite set is finite. Let $F\approx n$ with $n\in \mathbb{N}$ (Equinumerous sets, $A\approx B$ and $A⪯B$, The natural numbers $\mathbb{N}$ (von Neumann)), witnessed by a bijection $\phi :F\to n$, and let $B\subseteq F$. Then $\phi$ restricts to a bijection of $B$ onto $\phi [B]\subseteq n$ (Injection, surjection, bijection). Every element of the von Neumann natural $n$ is a natural number strictly smaller than $n$ (On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$), so $\phi [B]$ is a subset of $\mathbb{N}$ bounded above by $n$, hence finite by the sharper form of Every subset of an at most countable set is at most countable ("a subset $S\subseteq \mathbb{N}$ is finite if it is bounded above"). Since $\approx$ is symmetric and transitive, $B$ is finite.

(ii) A union of two finite sets is finite. First, if $H$ is finite and $g$ is any object then $H\cup \{g\}$ is finite: if $g\in H$ there is nothing to prove, and otherwise a bijection $u:H\to k$ extends to a bijection $H\cup \{g\}\to k\cup \{k\}=\sigma (k)$ by setting $u(g):=k$, which is injective because $k\mathrm{\notin }k$ (Every natural number is a transitive set and is not a member of itself). Now fix a finite set $F$ and argue by induction (The principle of mathematical induction) on $m\in \mathbb{N}$ over the statement "for every $G$ with $G\approx m$, the union $F\cup G$ is finite". At $m=0$ we have $G=\mathrm{\varnothing }$ and $F\cup G=F$. At $m=\sigma (j)$, a bijection $\psi :G\to \sigma (j)$ gives $g:={\psi }^{-1}(j)$ and $G^{\mathrm{\prime }}:=G\setminus \{g\}\approx j$ (restrict $\psi$), so $F\cup G=(F\cup G^{\mathrm{\prime }})\cup \{g\}$ is finite by the induction hypothesis and the previous sentence.

Discharge of the topology axioms.

Discrete. Every subset of $X$ lies in $\mathcal{P}(X)$, so (T1), (T2) and (T3) hold with nothing to check.

Indiscrete. (T1) is the definition. For (T2), a subfamily of $\{\mathrm{\varnothing },X\}$ has union $\mathrm{\varnothing }$ (if it is empty or $\{\mathrm{\varnothing }\}$) or $X$ (otherwise). For (T3), $\mathrm{\varnothing }\cap A=\mathrm{\varnothing }$ and $X\cap X=X$.

Cofinite. (T1): $\mathrm{\varnothing }$ is listed, and $X\setminus X=\mathrm{\varnothing }$ is finite. (T2): let $\mathcal{S}\subseteq {\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$. If every member is $\mathrm{\varnothing }$ the union is $\mathrm{\varnothing }$. Otherwise fix $U_0\in \mathcal{S}$ with $U_0\mathrm{\ne }\mathrm{\varnothing }$; then $X\setminus \bigcup \mathcal{S}\subseteq X\setminus U_0$, which is finite, so the left side is finite by (i). (T3): for nonempty $U,V$ with finite complements, $X\setminus (U\cap V)=(X\setminus U)\cup (X\setminus V)$ is finite by (ii); and if either of $U,V$ is empty so is $U\cap V$. The closed sets are the complements of the open ones, that is $X=X\setminus \mathrm{\varnothing }$ together with the finite sets.

Cocountable. Identical to the cofinite case with "at most countable" in place of "finite": (i) is replaced by Every subset of an at most countable set is at most countable itself, and (ii) by the statement that a union of two at most countable sets is at most countable, which is the two-set instance of Countable unions of at most countable sets, assuming ${\mathrm{A}\mathrm{C}}_{\omega }$ applied to the family $A_0:=U,A_1:=V,A_k:=\mathrm{\varnothing }$ for $k\ge 2$.

Particular point. (T1): $\mathrm{\varnothing }$ is listed and $p\in X$. (T2): a subfamily whose members are all $\mathrm{\varnothing }$ has union $\mathrm{\varnothing }$; otherwise some member contains $p$, hence so does the union. (T3): if $U$ and $V$ both contain $p$ then so does $U\cap V$; and if either is $\mathrm{\varnothing }$ then so is the intersection.

Sierpinski. The special case $X=\{a,b\}$, $p=b$ of the previous paragraph: the sets containing $b$ are $\{b\}$ and $S$, so ${\mathcal{T}}_b=\{\mathrm{\varnothing },\{b\},S\}={\mathcal{T}}_{\mathrm{S}\mathrm{i}\mathrm{e}\mathrm{r}}$.

• Two degenerate collapses. If $X$ is finite then the cofinite topology is the discrete one, since every subset then has finite complement by fact (i) above; if $X$ is at most countable the cocountable topology is discrete for the same reason. Both families are therefore interesting only on an infinite, respectively uncountable, set, and every statement made about them below names that hypothesis.

• Where the two extremes sit in the comparison order. The discrete topology is the finest and the indiscrete the coarsest topology on $X$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison): every topology is a subfamily of $\mathcal{P}(X)$ and contains $\mathrm{\varnothing }$ and $X$. Every other topology on $X$ lies between them, and the cofinite topology is coarser than the cocountable one, because a finite set is at most countable.

• No choice principle is needed for any of the six, despite the citation. The only appeal above that carries a choice hypothesis is Countable unions of at most countable sets, assuming ${\mathrm{A}\mathrm{C}}_{\omega }$, whose statement assumes ${\mathrm{A}\mathrm{C}}_{\omega }$, and it is used for a union of two sets only, padded with copies of $\mathrm{\varnothing }$. That instance is provable in ZF alone, by interleaving two given enumerations, exactly as The irrationals are uncountable records for the union of the rationals and the irrationals; the general theorem is cited because it is the form in which this library states the union result, not because the strength is needed. Nothing about the cocountable topology depends on countable choice.

• The Sierpinski point that is open is a genuine choice of labelling. Both $\{\mathrm{\varnothing },\{b\},S\}$ and $\{\mathrm{\varnothing },\{a\},S\}$ are topologies, and they are carried to each other by the transposition of $a$ and $b$; this library fixes the first and always names the open point.