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There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.

Use the order on locale maps in which $f\le g$ means $f^{\ast }(a)\le g^{\ast }(a)$ for every open $a$. A Hausdorff locale has a closed diagonal in its localic square. A totally unordered locale, or $T_U$ locale, has discretely ordered hom-sets: every comparable pair of maps with any common domain is equal.

In the localic square of $Y$, letThis is the largest open disjoint from the diagonal: inverse image along the diagonal sends the rectangle $a\times b$ to $a\wedge b$. Hence $N$ is the complement of its closure. If $f\le g$, thenfor every summand of $N$. Thus $(f,g)$ factors through the closure of the diagonal. For a Hausdorff locale that closure is the diagonal itself, so $f=g$. This proof of Hausdorff implies $T_U$ does not require classical logic.

Here is an explicit classical counterexample to the converse implication from spatial Hausdorffness. Give $\mathbb{R}$ the topology generated by the usual opens and by $\mathbb{Q}$. Every open has the form $U=V\cup (W\cap \mathbb{Q})$, where $V,W$ are usual open sets and we may arrange $V\subseteq W$. The topology is finer than the usual one, so the space is Hausdorff. Put $B=\mathrm{RO}(\mathbb{R})$, the complete Boolean algebra of ordinary regular open elements, and write $\rho (V)=\mathrm{int}\overline{V}$. Restriction from this space to its rational and irrational subspaces, followed by Booleanization, gives frame homomorphismsTo justify these formulas, each subspace has its ordinary subspace topology and is dense in the ordinary real line. Regular-open algebras of a space and a dense subspace are isomorphic: restrict a regular open and take its interior-closure, with inverse obtained by taking interior-closure in the ambient space. Restriction is a frame homomorphism, and double negation is a frame homomorphism into its fixed-element frame. These observations also show independence of the chosen representation of $U$. Since $V\subseteq W$, we have $f^{\ast }\le g^{\ast }$, but $f^{\ast }(\mathbb{Q})=0$ and $g^{\ast }(\mathbb{Q})=1$. Thus this Hausdorff space fails $T_U$ as a locale.

For the remaining arguments about connected locales, use the classical-logic option: positive opens of a locale are exactly the nonzero opens. There is a necessary positivity qualification. The empty locale has no positive opens, so the printed condition on those opens holds vacuously, although no point $1\to \varnothing$ exists. The point assertions therefore require $X$ to be positive. This qualification is automatic if “totally connected” includes nonemptiness; otherwise it must be added explicitly.

Let $X$ be positive and satisfy the condition on its positive opens of a locale. Put $F=\{a\in \mathcal{O}(X):a\ne 0\}$. It contains $1$, excludes $0$, and is upward closed. If $a,b\in F$ had $a\wedge b=0$, their union would be a positive open of a locale with a decomposition into two disjoint nonzero opens, contradicting its being a connected locale. Thus $a\wedge b\in F$. If ${\bigvee }_i{a}_i\ne 0$, not all $a_i$ can be zero; classical logic supplies a nonzero summand. Hence $F$ is a completely prime filter, and its characteristic frame homomorphism defines a point $p:1\to X$.

The point is dense. Indeed $p^{\ast }(a)=0$ implies $a=0$ by the definition of $F$. In nucleus language the image of $p$ has nucleus $p_{\ast }{p}^{\ast }$; its value at $0$ is $p_{\ast }(0)=0$, so the image is a dense sublocale.

Conversely suppose an open locale has a dense point $p$. Density means $p^{\ast }(a)=0$ only if $a=0$, and classically this implies $p^{\ast }(a)=1$ for every positive open of a locale. If a positive open $u$ split as $a\vee b=u$ with $a\wedge b=0$ and both summands nonzero, then $p^{\ast }(a)=p^{\ast }(b)=1$, contradicting preservation of their meet. Thus every positive open of a locale is a connected locale, proving the converse under the declared logical convention.

Let $x:X\to 1$ be the terminal map. For a positive totally connected locale $X$, the point above satisfiesbecause the right side is $1$ when $a\ne 0$ and both sides are $0$ otherwise. Hence $1_X\le px$. If $f:X\to Y$ and $Y$ is a totally unordered locale, composition gives $f\le fpx$, so $f=fpx$. Thereforewhich is the required constant factorization. For the empty source the factorization need not exist: the unique map $\varnothing \to \varnothing$ does not factor through $1$. Thus the positivity qualification matters for this assertion too.

For a sequence $ϵ$, let $C_n(ϵ)$ be its prefix cylinder set, consisting of sequences agreeing with $ϵ$ in the first $n$ coordinates. The specified collection consists exactly of unions of such cylinder sets.

The empty set meets the defining condition vacuously, and $X$ meets it because every cylinder set is contained in $X$. For a union ${\bigcup }_{\lambda }{Y}_{\lambda }$, a point lies in one $Y_{\lambda }$ and has a prefix cylinder set contained in that set, hence contained in the union. For a finite intersection, a point has a cylinder neighbourhood in every constituent. Taking the largest of the finitely many prefix lengths gives one cylinder set contained in all of them. Therefore the collection contains the empty and whole sets, is closed under arbitrary unions and finite intersections, and is a topology on $X$, the prefix topology of Cantor space.

There is a small defect in the printed definition: the listed family omits $\mathbb{R}$ and is not a topology, since ${\bigcup }_{n\ge 1}(-n,\infty )=\mathbb{R}$. We use the intended upper-ray topology on the real line, obtained by adjoining $\mathbb{R}$ to that family.

Take $f(n)=n$. For every real $a$,has finite complement and is therefore open in the cofinite topology. The inverse images of the empty set and the whole line are also open. Hence this nonconstant function is a continuous map for the corrected upper-ray topology on the real line. The same inverse-image calculation verifies the condition for every ray literally listed in the question.

Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, with each subset identified with its increasing enumeration. In the homogeneous-cone convention, a family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense when some infinite $M$ satisfiesThe stronger Ramsey set of infinite subsets convention asks for such an infinite $M\subseteq A$ inside every given infinite ground set $A$. We will prove the open-set assertion in this stronger form, and our Ramsey-but-not-completely-Ramsey example will satisfy it as well. Thus the distinction in conventions does not affect any of the conclusions below.

For a non-Ramsey example, declare $X\sim Y$ if their symmetric difference is finite, and choose a representative $R$ from each equivalence class using the axiom of choice. Setfor the representative of the class of $X$. Removing one point reverses this colour, because its symmetric difference with $R$ changes by exactly one point. Consequently every cone $[M{]}^{\omega }$ contains the oppositely coloured sets $M$ and $M\setminus \{\min M\}$. Either colour class of this finite-symmetric-difference parity colouring is not a Ramsey family in the homogeneous-cone sense. No definability is claimed for the representative selection.

The ordinary topology on infinite subsets, denoted $\tau$, has basic cylinders $[s]=\{X:s⊏X\}$, where $s$ is a finite stem of the increasing enumeration. Let $O$ be a $\tau$-open family and fix an arbitrary infinite ground set $A$. We give a complete fusion proof of a homogeneous cone inside $A$.

Use the convention $\max \varnothing =0$. For finite $s$ and an infinite reservoir $B$ above $\max s$, say that $B$ accepts $s$ if $[s,B]\subseteq O$, whereSay that $B$ rejects $s$ if no infinite subset of $B$ accepts $s$. There is always a refinement deciding $s$: choose an accepting subreservoir if one exists, and otherwise retain $B$, which rejects. Both decisions persist under further thinning.

A first fusion produces an infinite $H\subseteq A$ deciding every finite $s\subseteq H$ on its tail $H/s=\{h\in H:h>\max s\}$. First decide the empty stem. Choose the first point, and thin the remaining reservoir successively to decide every subset of the selected prefix. Repeat after choosing each new point. There are finitely many subsets to decide at each stage. The diagonal sequence of chosen points has the required property, since its tail after the largest point of any stem lies in the reservoir where that stem was decided. The empty-stem decision persists too. This is deciding all finite stems by fusion.

If $H$ accepts the empty stem, $[H{]}^{\omega }\subseteq O$ and we are done. Suppose it rejects. For a rejected finite $s\subseteq H$, only finitely many $h\in H/s$ can have $H/h$ accepting $s\cup \{h\}$. Otherwise let $D$ be their infinite set. Every infinite $X\subseteq D$ starts with some such $h$, and its remaining tail lies in $H/h$, so $s\cup X\in O$. Then $D$ accepts $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

A second fusion constructs $K\subseteq H$ whose every finite subset is rejected. If a finite prefix of chosen points has been selected, all its subsets are rejected by induction. Avoid the union of their finitely many accepting successor sets when choosing the next point. The first fusion already decides every successor, so every new subset is rejected. An infinite reservoir remains after each finite exclusion.

If some $X\in [K{]}^{\omega }$ belonged to $O$, openness would give an initial segment $s⊏X$ with $[s]\subseteq O$. Its reservoir $H/s$ would then accept $s$, contradicting the second fusion. Therefore $[K{]}^{\omega }\cap O=\varnothing$. In either case we have a homogeneous cone inside the arbitrary ground set $A$. Every $\tau$-open family is a Ramsey set of infinite subsets, under both conventions.

A completely Ramsey set requires more: for every finite $s$ and infinite reservoir $A$ above $\max s$, there is an infinite $B\subseteq A$ such thatThe finite stem must be kept fixed. These $[s,A]$ are the basic neighbourhoods of the Ellentuck topology, also called the star topology.

Let $E$ be the even positive integers, and construct the finite-symmetric-difference parity colouring $\epsilon$ on $[E{]}^{\omega }$. DefineFor any infinite ground set $A$, the set $A\setminus \{1\}$ is still infinite and its whole cone misses $\mathcal{R}$. Thus $\mathcal{R}$ is a Ramsey set of infinite subsets even in the stronger convention. But inside $[\{1\},E]$, every infinite refinement $B\subseteq E$ gives the two sets $\{1\}\cup B$ and $\{1\}\cup (B\setminus \{\min B\})$ of opposite colours. Hence no such refinement is homogeneous. The family $\mathcal{R}$ is Ramsey but not completely Ramsey. The stem-supported Ramsey family need not be completely Ramsey construction explains this example: forgetting the stem $\{1\}$ is precisely what loses the obstruction.

Finally take $D=[E{]}^{\omega }$. It is nonempty and star-open, since it equals $[\varnothing ,E]$. It is $\tau$-closed: if an infinite set contains an odd integer, an initial segment witnessing that integer has a cylinder disjoint from $D$. Its $\tau$-interior is empty: after any finite stem of even integers, append an arbitrarily large odd integer and then infinitely many further points. ThusThis is the cone on an infinite coinfinite ground set. Choose its subset $Y=\{X\in [E{]}^{\omega }:\epsilon (X)=0\}$. Since ${\overline{Y}}^{\tau }\subseteq D$, $Y$ is also $\tau$-nowhere dense. Nevertheless every $[\varnothing ,B]$ with infinite $B\subseteq E$ contains both $B$ and $B\setminus \{\min B\}$, of opposite colours. Hence $Y$ is not completely Ramsey. We have proved explicitly that ordinarily nowhere-dense sets need not be completely Ramsey; the topology in the nowhere-dense hypothesis cannot be silently changed to the star topology.

This proof uses neither the axiom of choice nor a preliminary choice of points in all the given sets. For each nonempty $A_j$, form a disjoint union $X_j=A_j\bigsqcup \{{\infty }_j\}$ with a distinguished added point, and give it the topologyThe original set $A_j$ has the indiscrete subspace topology, and is clopen. Any open cover of $X_j$ either contains $X_j$, or contains both $A_j$ and $\{{\infty }_j\}$, so it has a subcover of at most two members. Thus $X_j$ is compact by a proof requiring no choices. No Hausdorff restriction is imposed in this question, and these generally non-Hausdorff spaces are allowed.

The product $X={\prod }_{j\ge 1}{X}_j$ has the explicitly given all-infinity point, so it is nonempty without choice. By the assumed countable-product compactness it is compact. For each $j$ letThe coordinate projection is continuous and $A_j$ is closed in $X_j$, so $F_j$ is closed. Any finite intersection of the $F_j$ is nonempty: pick an element of each of those finitely many nonempty sets and put ${\infty }_k$ in every other coordinate. Finite choice follows by induction in ordinary set theory without assuming the axiom of choice.

Compactness and the finite intersection property now give ${\bigcap }_{j\ge 1}{F}_j\ne \varnothing$. Choose one point $x$ from this single nonempty set. The function $f(j)=x_j$ satisfies $f(j)\in A_j$ for every $j$, proving the axiom of countable choice.

The implication holds in choice-free set theory: countable product compactness implies countable choice. The printed lower index zero is inconsistent with the stated positive-integer index set; indexing from one merely relabels this argument and changes no conclusion.

Include the empty set, and take all unions of intervals $[a,b)$ with $a<b$. These intervals cover the real line, and the intersection of two is either empty or $[\max (a,c),\min (b,d))$. Distributing intersections over unions therefore makes finite intersections of open sets open; arbitrary unions are open by construction. The whole line is open, so the axioms of a topology hold. This is the lower limit topology, whose space is the Sorgenfrey line.

If $x<y$, the neighborhoods $[x,y)$ and $[y,y+1)$ are disjoint. Hence $\boxed{(\mathbb{R},{\tau }_1)\text{ is Hausdorff}}$.

For the identity into the cofinite topology, an inverse image is the same subset of $\mathbb{R}$. Let $F$ be finite and $x\notin F$. If $F$ has a point greater than $x$, choose $b$ to be its least such point; otherwise take $b=x+1$. Then $[x,b)$ avoids every point of $F$, including $b$ itself. Taking these intervals over all $x\notin F$ writes $\mathbb{R}\setminus F$ as a ${\tau }_1$-open set. The inverse image of every cofinite open set, and of the empty set, is open. The identity map is continuous, because ${\tau }_2\subseteq {\tau }_1$.

Use the separated-set definition of topological entropy. For a compatible metric $d$ and $n\ge 1$, define the Bowen metric and separated-set count byFor fixed $n$, the metric $d_n$ generates the original topology: it includes the $k=0$ term and all finitely many iterations of a map are continuous. Thus its space is compact and $s_n(d,ϵ)$ is finite. With natural logarithms, setThe last limit exists, possibly with value infinity, because decreasing $ϵ$ can only increase separated-set counts.

Let $d^{\prime }$ be another compatible metric. The identity $(X,d)\to (X,d^{\prime })$ is a continuous function on a compact domain, hence uniformly continuous. For every $ϵ>0$, choose $\delta >0$ such that $d(x,y)<\delta$ implies $d^{\prime }(x,y)<ϵ$. The same implication at every iterate givesBy its contrapositive, every $ϵ$-separated set for $d_n^{\prime }$ is $\delta$-separated for $d_n$. Consequently $s_n(d^{\prime },ϵ)\le s_n(d,\delta )$ for every $n$, andTaking $ϵ\downarrow 0$ gives $h_{d^{\prime }}(f)\le h_d(f)$. Interchanging the two compatible metrics gives the reverse inequality. ThereforeCompactness is used for uniform continuity, not merely for pointwise equivalence of the two metrics.

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit natural transformation $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$, satisfyingAn algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aTa=a{\mu }_A$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ satisfies $ha=bTh$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:U$ with $F⊣U$, the monad induced by an adjunction is $T=UF$, with $\mu =U\epsilon F$. The Eilenberg-Moore comparison functor is $K(D)=(UD,U{\epsilon }_D)$ and $K(h)=Uh$. The functor $U$ is monadic when this comparison is an equivalence of categories; this is a monadic adjunction.

The precise monadicity theorem says that $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, reflects isomorphisms, and satisfies the following condition: whenever the image of a parallel pair in $\mathcal{D}$ admits a split coequalizer in $\mathcal{C}$, the pair has a coequalizer in $\mathcal{D}$ and $U$ preserves that coequalizer. Equivalently, these coequalizers are created up to the canonical isomorphism identifying their underlying coequalizer objects. Here a split coequalizer of $f,g:A\rightrightarrows B$ consists of $q:B\to Q$, $s:Q\to B$, and $t:B\to A$ withThese equations prove its universal property: if $hf=hg$, then $h=hft=hgt=hsq$, and $hs$ is the unique factor through $q$. Every functor preserves these equations, so every functor preserves split coequalizers.

First consider the forgetful functor $U^T:{\mathcal{C}}^T\to \mathcal{C}$. It has the free algebra functor $A\mapsto (TA,{\mu }_A)$ as a left adjoint: the inverse transposition formulas are $h\mapsto h{\eta }_A$ and $v\mapsto bTv$. An monad algebra morphism whose underlying arrow is invertible has an algebra-morphism inverse, by rearranging its structure equation. Thus $U^T$ reflects isomorphisms.

Suppose $f,g:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$. Applying $T$ gives a coequalizer of $Tf,Tg$. The equation $qbTf=qfa=qga=qbTg$ therefore gives a unique $c:TQ\to Q$ with $cTq=qb$. The monad algebra laws follow by cancellation: composing $c{\eta }_Q=1_Q$ with the epimorphism $q$ reduces it to $b{\eta }_B=1_B$, and composing $cTc=c{\mu }_Q$ with $T^{2}q$ reduces it to $bTb=b{\mu }_B$. Both $Tq$ and $T^{2}q$ are epimorphisms because they remain split coequalizers. Thus $(Q,c)$ is an algebra for a monad and $q$ is an monad algebra morphism. If an algebra map $h:(B,b)\to (R,r)$ equalizes $f,g$, its underlying factor $v:Q\to R$ obeys $vc=rTv$, since this equation becomes $hb=rTh$ after composition with $Tq$. This proves that $U^T$ creates the required coequalizers. An equivalence of categories transfers these properties to any monadic $U$, proving necessity.

Conversely assume the three conditions and write $T=UF$. Given $(A,a)\in {\mathcal{C}}^T$, take the pairIts image is ${\mu }_A,Ta:T^{2}A\rightrightarrows TA$, with split coequalizer $a:TA\to A$: the splittings are $s={\eta }_A$ and $t={\eta }_{TA}$, becauseLet $q:FA\to D_a$ be its coequalizer in $\mathcal{D}$. Preservation gives a canonical isomorphism $U{D}_a\cong A$ under which $Uq=a$. Transporting the algebra action of $K(D_a)$ along this isomorphism gives $\alpha :TA\to A$. Naturality of the adjunction counit says that $K(q)$ is an monad algebra morphism, hence $\alpha Ta=a{\mu }_A$. But $aTa=a{\mu }_A$ too, and $Ta$ is a split epimorphism, with section $T{\eta }_A$. Therefore $\alpha =a$. Every algebra is consequently isomorphic to some $K(D_a)$.

It remains to establish full faithfulness, rather than merely construct objects. For $B\in \mathcal{D}$, put $b=U{\epsilon }_B$. The adjunction counit ${\epsilon }_B:FUB\to B$ coequalizes ${\epsilon }_{FUB},Fb$. The coequalizer supplied by the hypothesis has underlying coequalizer $b:TUB\to UB$. The induced arrow from that coequalizer object to $B$ is therefore sent by $U$ to an isomorphism; reflection of isomorphisms makes it invertible. Thus ${\epsilon }_B$ itself is this coequalizer.

Now let $h:K(B)\to K(C)$ be an monad algebra morphism, and put $c=U{\epsilon }_C$. The arrow $v={\epsilon }_{C}Fh:FUB\to C$ coequalizes the two arrows in the preceding presentation. To check this directly, use transposition along $F⊣U$: the transposes of $vFb$ and $v{\epsilon }_{FUB}$ are respectively $hb$ and $cTh$, which agree by the algebra-morphism equation. Hence a unique $\bar{h}:B\to C$ satisfies $\bar{h}{\epsilon }_B=v$. Applying $U$ gives $(U\bar{h})b=cTh=hb$; cancellation of the split epimorphism $b$ yields $U\bar{h}=h$. Finally, any arrow $r:B\to C$ satisfies $r{\epsilon }_B={\epsilon }_{C}FUr$ by naturality, so two arrows with equal image under $U$ are equal by the coequalizer property. The comparison is full and faithful and essentially surjective, hence an equivalence of categories. This proves the theorem.

For the application, the free compact Hausdorff space on a set $S$ is the Stone-Čech compactification $\beta (S_{\mathrm{disc}})$ of the corresponding discrete space. Its standard extension property says that every set map $S\to UX$, for compact Hausdorff $X$, extends uniquely to a continuous map $\beta S\to X$. This constructs the required left adjoint. Also a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so $U$ reflects isomorphisms.

Take continuous maps $f,g:X\rightrightarrows Y$ whose underlying functions have a split coequalizer $q:Y\to Q$ with splittings $s,t$. The splittings need not be continuous. LetThe set $E$ is compact and hence closed. Since $qf=qg$, it is contained in $R$. Conversely $(y,sq(y))=(ft(y),gt(y))\in E$. ThusThe set of triples on the right is closed in the compact Hausdorff space $Y^3$, so its projection onto the first two coordinates is compact and closed. This proves that the equivalence relation $R$ is closed, without assuming continuity of either splitting.

Give $Q\cong Y/R$ the quotient topology. The allowed closed-relation criterion makes it a compact Hausdorff space. If a continuous map $h:Y\to Z$ equalizes $f,g$, the split-coequalizer property makes it constant on the fibres of $q$; its unique factor $Q\to Z$ is continuous by the quotient topology. Hence this is a coequalizer in compact Hausdorff spaces, preserved by $U$. Its topology is forced uniquely, since any continuous surjection from a compact space to a Hausdorff space is a quotient map. All three conditions hold, and therefore the forgetful functor from compact Hausdorff spaces to sets is monadic.

For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, write its adjunction unit as $\eta$ and its adjunction counit as $\epsilon$. It is an idempotent adjunction when the induced monad has invertible multiplication $\mu =R\epsilon L:RLRL\Rightarrow RL$. The dual condition is invertibility of the induced comonad comultiplication $\delta =L\eta R:LR\Rightarrow LRLR$.

To prove equivalence of these conditions, suppose $\mu$ is invertible. The two unit identities imply $T\eta =\eta T={\mu }^{-1}$, where $T=RL$. For any algebra for a monad $a:TA\to A$, naturality giveswhile $a{\eta }_A=1_A$. Thus ${\eta }_A$ is invertible. For each $D\in \mathcal{D}$, the arrow $R{\epsilon }_D:T(RD)\to RD$ is a monad algebra action, so ${\eta }_{RD}$ is invertible. Applying $L$ proves ${\delta }_D=L{\eta }_{RD}$ invertible. Apply this implication to the opposite adjunction, interchanging the roles of the monad and comonad, to obtain the converse. Therefore idempotence is self-dual.

The open-set frame $O(X)$ is ordered by inclusion. Its arbitrary joins are unions; arbitrary meets are interiors of intersections. Finite intersections are already open, so finite meets are ordinary intersections, including the empty meet $X$. Distributivity of intersection over arbitrary unions proves the frame law. If $f:X\to Y$ is a continuous map, inverse imagepreserves arbitrary unions and finite intersections, including $\varnothing$ and the entire space. It is therefore a frame homomorphism. Since $(gf{)}^{-1}=f^{-1}{g}^{-1}$ and $1_X^{-1}=1_{O(X)}$, this defines $O:\mathbf{Top}\to {\mathbf{Frm}}^{\mathrm{op}}$.

Let $2=\{0<1\}$ and let $P(A)$ be the set of points of a frame $A$, namely frame homomorphisms $A\to 2$. For $a\in A$, put $U_a=\{p:p(a)=1\}$. Preservation of joins and finite meets givesIn the last equality, a join in $2$ is $1$ precisely when at least one summand is $1$, also covering the empty family. Thus the sets $U_a$ are all the opens of a topology, rather than merely a proposed basis.

For a frame homomorphism $h:A\to B$, define $P(h):P(B)\to P(A)$ by $p\mapsto p\circ h$. Since $P(h{)}^{-1}(U_a)=U_{h(a)}$, it is continuous. Composition and identities follow from composition of functions, giving $P:{\mathbf{Frm}}^{\mathrm{op}}\to \mathbf{Top}$.

The frame-point adjunction is the natural bijectionA continuous map $f:X\to P(A)$ determines $h_f(a)=f^{-1}(U_a)$; the displayed identities for $U_a$ show that $h_f$ is a frame homomorphism. Conversely a frame homomorphism $h:A\to O(X)$ determines $f_h(x)(a)=1$ exactly when $x\in h(a)$. Membership in unions and finite intersections proves that $f_h(x)$ is a frame homomorphism to $2$, and $f_h^{-1}(U_a)=h(a)$ proves continuity. These constructions are inverse. Precomposing in $X$ becomes inverse image of open sets; precomposing a frame homomorphism becomes composition of frame points. Hence the bijection is natural in both arguments and proves $O⊣P$.

The adjunction unit isThe adjunction counit at $A$, regarded as an arrow of ${\mathbf{Frm}}^{\mathrm{op}}$, is represented by the frame homomorphism $e_A:A\to O(P(A))$, $a\mapsto U_a$. It is surjective because every open of $P(A)$ is of this form. We can compute $P(e_A):P(O(P(A)))\to P(A)$ explicitly: it sends $q$ to $p=q\circ e_A$. For every open $U_a$ of $P(A)$,Thus ${\eta }_{P(A)}P(e_A)=1$. Conversely, evaluation at $p$ sends $U_a$ to $p(a)$, so $P(e_A){\eta }_{P(A)}=1$. Both maps are continuous by their definitions. Therefore $P(e_A)$ is a homeomorphism. The multiplication of the induced monad $PO$ at $X$ is $P(e_{O(X)})$, which is invertible by this calculation. Consequently the frame-point adjunction is idempotent. This does not assert that every ${\eta }_X$ is a homeomorphism: it says that applying the point-space construction a second time makes no further change.

This does not define a topology on $\mathbb{Z}$. Every singleton $\{2k\}$ is declared open, but their union over $k\in \mathbb{Z}$ is $2\mathbb{Z}$, an infinite proper subset. It is therefore not declared open, violating closure under arbitrary unions. This is the example that finite subsets do not define a topology on an infinite set. The empty set, whole space and finite-intersection requirements do hold, but they do not compensate for the failed union axiom. This is different from the cofinite topology, which uses finite complements rather than finite sets.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.

False. Give $X=\mathbb{R}$ its usual topology, let $Y$ be a one-point space, and take the constant continuous map $f:X\to Y$. The set $K=Y$ is compact, but $f^{-1}(K)=\mathbb{R}$ is not compact. For example, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. Thus continuity alone does not preserve compactness under inverse images.

Every nonempty open set in the given topology contains $1$. Suppose a continuous map $f$ took distinct values $f(m)$ and $f(n)$. Choose disjoint open intervals $U,V\subset \mathbb{R}$ containing these respective values. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty open sets in $\mathbb{N}$, and therefore both contain $1$. This is impossible because the inverse images of disjoint sets are disjoint. Thus every such continuous map is constant. Conversely, constant maps are continuous. This is the continuous real maps from the initial-segment topology argument, using the Hausdorff separation of distinct real points.

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A profinite group has either of the following equivalent descriptions: a compact Hausdorff totally disconnected topological group, or an inverse limit of finite discrete groups. Totally disconnected means that every connected component is a singleton.

For clarity, an inverse system over a directed set $I$ consists of groups $G_i$ and maps ${\varphi }_{ij}:G_j\to G_i$ for $i\le j$, with ${\varphi }_{ii}=1$ and ${\varphi }_{ik}={\varphi }_{ij}{\varphi }_{jk}$. Its inverse limit iswith coordinatewise multiplication and the subspace topology. Equivalently, it is a group with compatible continuous projections such that every other compatible family of maps into the $G_i$ factors uniquely through it.

The displayed limit is a closed subgroup of the compact Hausdorff product of the finite groups, and is totally disconnected, since two different tuples are separated by a clopen coordinate condition. Conversely a compact Hausdorff totally disconnected group has a neighbourhood basis of open normal subgroups. One way to see the group part of this standard clopen-basis fact is to choose a clopen identity neighbourhood $U$ inside a prescribed neighbourhood. Compactness of $U$ and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$. The stabilizer $H=\{g:gU=U\}$ contains $V$, is open, and is contained in $U$ because $1\in U$. Its index is finite by part (ii); its normal core is the intersection of finitely many conjugates and is open and normal. The canonical mapis injective because these $N$ separate points, and surjective because compatible cosets have the finite intersection property and $G$ is compact. It is a homeomorphism by compactness and the Hausdorff property. This explains the equivalence of the two definitions.

The profinite Frattini subgroup is $\Phi (G)=\bigcap M$, where $M$ ranges over all maximal proper open subgroups; for the trivial group the empty intersection is $G$. For a pro-p group,Here powers and group commutators mean generated subgroups and the bar supplies closure. Every maximal open subgroup in a pro-p group is normal of index $p$, as follows by passing to the finite p-group quotient by its normal core. Such a subgroup therefore contains all powers and group commutators. Conversely the quotient by $D=\overline{G^p[G,G]}$ is an abelian pro-p group of exponent $p$. If an element of this quotient is nontrivial, some finite elementary abelian quotient detects it; a linear functional to ${\mathbb{F}}_p$ then detects it as well. Its group homomorphism kernel pulls back to a maximal open subgroup of $G$. Thus the intersection of these kernels is exactly $D$.

We also use the topological generation criterion: a subset generates $G$ topologically if and only if its image generates $G/\Phi (G)$ topologically. To prove the nontrivial direction, a proper closed generated subgroup $K$ lies in a proper open subgroup $KN$ for a suitably small open normal $N$, and hence in a maximal open subgroup obtained from the finite quotient $G/N$. Since that maximal subgroup contains $\Phi (G)$, the image of $K$ cannot generate the Frattini quotient.

Now suppose $G$ is a nontrivial procyclic pro-p group. If it had distinct maximal open subgroups $M_1,M_2$, the homomorphism to $C_p\times C_p$ induced by the two quotient maps would be onto: its projections are onto, and a proper such subgroup would be a one-dimensional graph, forcing the two kernels to coincide. Hence $G/(M_1\cap M_2)\cong C_p^2$, contradicting procyclicity. Thus there is exactly one maximal open subgroup $M$, and $\Phi (G)=M$. Any $g\notin M$ generates $G/\Phi (G)\cong C_p$, so the criterion givesFor the trivial group use its identity as generator. The additive group ${\mathbb{Z}}_p$ is an infinite example, topologically generated by $1$: its finite continuous quotients are cyclic $\mathbb{Z}/p^n\mathbb{Z}$.

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.