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Every subset of a discrete space is open, so its Borel sigma-algebra is the full power set.

In the Stone-Čech compactification of a discrete space, the closure of each subset of $X$ is clopen, because its indicator extends to a continuous zero-one-valued function. For disjoint closed sets $A,B\subseteq \beta X$, choose $u=0$ on $A$ and $u=1$ on $B$ by Urysohn's lemma. Then $U=\overline{\{x\in X:u(x)<1/2\}}$ and $V=\beta X\setminus U$ are disjoint clopen sets containing $A$ and $B$, respectively. Density proves these containments.

Let $K\subseteq X$, where the ambient topological space $X$ is Hausdorff and $K$ is compact in its subspace topology. Then $K$ is a closed set in $X$: the complement $X\setminus K$ is an open set in $X$. The ambient space need not be compact. For example, $[0,1]$ is a compact closed interval in the non-compact Hausdorff real line.^[ref]

The ambient Hausdorff hypothesis matters even when $K$ itself is Hausdorff. In the Sierpiński space $S=\{0,1\}$, the open sets are $\varnothing ,\{1\},S$, so the closed sets are $\varnothing ,\{0\},S$. Thus $K=\{1\}$ is compact and Hausdorff in its one-point subspace topology, but is not closed in $S$: its complement $\{0\}$ is not an open set. Every open neighborhood of $0$ contains $1$, so separation from $K$ fails. In contrast, the discrete topology on the same underlying set is Hausdorff and makes both singletons closed.

The diagram lists every open set and closed set, so the distinction can be checked directly. All subsets in both panels are compact: for any open cover of a finite set, choose one covering open set per point, obtaining a finite subcover. Compactness alone therefore cannot distinguish the two panels.

Here is the finite-selection mechanism in the general proof. For $x\in X\setminus K$ and each $k\in K$, the Hausdorff property gives disjoint open neighborhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ give an ambient open cover of $K$. Select a finite subcover $U_{k_1},\dots ,U_{k_m}$ and setThen $K\subseteq U$, $x\in V$, and $U\cap V=\varnothing$. The finite intersection makes $V$ an open set; an infinite intersection would not guarantee this. Thus every point of $X\setminus K$ has an open neighborhood contained in that complement, proving that $K$ is closed. The empty set is already closed.

This theorem combines with the continuous image of a compact space theorem to make every continuous map from a compact space to a Hausdorff space a closed map: a closed subset of its domain is compact, and its image is compact and hence closed in the target. This is the reason the inverse in the compact-to-Hausdorff continuous bijection theorem is continuous. The converse implication, from closed to compact, needs additional hypotheses: the real line is closed in itself but not compact; in a compact space, every closed subset is compact.

For a discrete space $X$ and $D\subseteq X$, the continuous extension of $1_D$ to $\beta X$ remains idempotent by density, so it takes values in $\{0,1\}$. Its one-set is the clopen closure of $D$. If $O$ is any open set in $\beta X$, density gives $\overline{O}=\overline{O\cap X}$, so this closure is clopen as well. Thus $\beta X$ is an extremally disconnected space. For infinite $X$, this does not imply that every subset of $\beta X$ has open closure.

Local time records how much a stochastic process visits a level. Its normalization depends on the process: when the state space is a discrete space, it is occupation time at a state, while local time of a semimartingale is an occupation density weighted by quadratic variation. Brownian local time is a special case.

For the model $F$ of the Möbius band, the tangent vectors along its core circle of a Möbius band areTheir cross product, divided by its norm $R$, gives the continuous unit normalAlthough $F(2\pi ,0)=F(0,0)$, we have $n(2\pi )=-n(0)$. Any other continuous choice of unit normal along this parametrized interval is either $n$ everywhere or $-n$ everywhere: its relative sign is a continuous map into the discrete space $\{+1,-1\}$, and the interval is connected. Thus no choice can agree at the identified endpoints. This short calculation proves that the Möbius band has no global continuous unit normal and is not an orientable surface.

The plotted quantity is $n(\theta )\cdot n(0)=\cos (\theta /2)$. It compares ambient vectors at generally different points; after one circuit it is $-1$ at the starting point, and after two circuits it is $+1$ there. Only these endpoint comparisons are needed for the orientability argument.

The extension. Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal topological space can be separated by a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$.

For the bounded Tietze extension theorem, first suppose $g:C\to [-M,M]$ is continuous. The subsets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are closed in $X$, because $C$ is closed. Urysohn's lemma supplies a continuous function $h:X\to [-M/3,M/3]$ with $h=-M/3$ on $A$ and $h=M/3$ on $B$. Directly checking the two outer ranges and the middle range of $g$ givesEmpty level sets cause no difficulty: take the appropriate constant, or zero if both are empty. This is the Urysohn two-thirds approximation.

Apply this successively to the residual on $C$, beginning with $f$ and $M_0=1$. We get $h_n$ with $\Vert h_n{\Vert }_{\infty }\le \frac{1}{3}(\frac{2}{3}{)}^n$ and residual after the first $n$ terms at most $(2/3{)}^n$. The Weierstrass M-test gives uniform convergence of $F={\sum }_{n\ge 0}{h}_n$. By the uniform limit theorem, $F$ is continuous; the residual bound proves $F{|}_C=f$, and the sum of the norm bounds is one. Thus

Closed sets as exact zero sets. If $X$ is a perfectly normal space, write a closed $C$ as ${\bigcap }_{n\ge 0}{U}_n$ with each $U_n$ open. Apply Urysohn's lemma to $C$ and $X\setminus U_n$ to obtain $f_n:X\to [0,1]$ vanishing on $C$ and equal to one on $X\setminus U_n$. The uniformly convergent seriesIndeed all summands vanish on $C$; outside $C$, at least one summand is strictly positive. The uniform limit theorem proves continuity. This realizes the closed G-delta set as a zero set construction.

Conversely, suppose each closed set is the zero set of a continuous function. For disjoint closed $C,D$, select $f,g:X\to [0,1]$ with zero sets $C,D$. Their sum never vanishes, andThe quotient is continuous. This also proves normality by inverse images of disjoint endpoint neighborhoods. Each zero set is a G-delta set, since $f^{-1}(0)={\bigcap }_{n\ge 1}{f}^{-1}([0,1/n))$; thus this property characterizes perfectly normal spaces among T1 spaces.

Subspaces. Let $Y\subseteq X$ and let $A,B$ be disjoint closed subsets of $Y$. Put $C={\overline{A}}^X$ and $D={\overline{B}}^X$. Since $C\cap Y=A$ and $D\cap Y=B$, the intersection $C\cap D$ misses $Y$. Choose continuous functions $f,g$ on $X$ whose zero sets are $C,D$. On the open neighborhood $W=X\setminus (C\cap D)$ of $Y$, the same quotient $f/(f+g)$ is continuous and is zero on $A$, one on $B$. Its inverse images of $[0,1/3)$ and $(2/3,1]$, restricted to $Y$, are disjoint open neighborhoods. Therefore every subspace is a normal topological space: perfect normality implies hereditary normality.

A strict counterexample to the converse. Let $D$ be an uncountable discrete space, and let $X=D\cup \{\infty \}$ be its one-point compactification. Its neighborhoods of $\infty$ have finite complement in $D$, because compact subsets of a discrete space are finite. The space is compact Hausdorff, hence a T1 space and normal. A subspace missing $\infty$ is discrete. A subspace containing $\infty$ is closed in $X$, since its complement is a subset of the open discrete part; it is therefore compact Hausdorff and normal. Thus all subspaces are normal.

But $\{\infty \}$ is not a G-delta set. A countable intersection of its open neighborhoods omits at most countably many elements of $D$, and cannot omit all of uncountable $D$. Hence $X$ is not a perfectly normal space. This is the uncountable discrete one-point compactification counterexample.

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.

A Hausdorff compact topology. The empty set omits zero and $\mathbb{Z}$ has finite complement. An arbitrary union omitting zero belongs to the family. If a union contains zero, one of its constituent sets contains zero and is cofinite, so the union is also cofinite. A finite intersection either omits zero or is an intersection of finitely many cofinite sets and is therefore cofinite. This proves all topology axioms.

For distinct nonzero integers $a,b$, the disjoint open singletons $\{a\},\{b\}$ separate them. To separate $0$ and $n\ne 0$, use the disjoint open sets $\mathbb{Z}\setminus \{n\}$ and $\{n\}$. Thus this is a Hausdorff space. Every open cover has a member containing zero, which leaves only finitely many integers uncovered; choose one member for each of those points. This gives a finite subcover and proves compactness. The space is the one-point compactification of the discrete space $\mathbb{Z}\setminus \{0\}$, with zero as the added point.

False. The rational numbers $\mathbb{Q}$ with their subspace topology from $\mathbb{R}$ are Hausdorff. If a subset contains two distinct rationals $a<b$, choose an irrational $c$ between them; intersecting the subset with $(-\infty ,c)$ and $(c,\infty )$ gives a separation into two nonempty relatively open sets. Thus every nonempty connected subset of $\mathbb{Q}$ is a singleton, so it is totally disconnected. Yet no rational is isolated, because every interval around it contains another rational. Therefore $\mathbb{Q}$ is not a discrete space. Hausdorffness and total disconnectedness do not imply discreteness.

A completely regular Hausdorff space is a Hausdorff space in which each point $x$ outside a closed set $F$ is separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. Put $E=C_b(X)$ and define the evaluation character by ${\delta }_x(g)=g(x)$. The weak-star topology on $E^{\prime }$ is the topology of pointwise convergence on $E$: its coordinate maps $p\mapsto p(g)$ are continuous, and generate the topology. It is Hausdorff, and the dual unit ball is compact in it by the Banach-Alaoglu theorem.

For nonempty $X$, $\Vert {\delta }_x\Vert =1$ by the supremum norm bound and the constant function $1$. All the coordinates of $x\mapsto {\delta }_x$ are continuous, so $\delta$ is continuous. Complete regularity separates distinct points, proving injectivity. To prove continuity of the inverse onto its image, let $x\in O$ with $O$ open in $X$. Separating $x$ from $X\setminus O$ gives a bounded continuous $u$ withSuch neighborhoods prove that $\delta$ is a homeomorphism onto $\delta (X)$. This qualification is essential: it is not onto the whole dual, since the zero functional is not an evaluation on a nonempty space.

Define the Stone-Čech compactification to beThe Banach-Alaoglu theorem makes this a compact Hausdorff space, with $X$ identified with its dense evaluation copy. For $g\in C_b(X)$, the coordinate function $\tilde{g}(p)=p(g)$ is a continuous extension. Density gives $\Vert \tilde{g}{\Vert }_{\infty }=\Vert g{\Vert }_{\infty }$. Conversely any $h\in C(\beta X)$ restricts to an element $g$ of $C_b(X)$, and $h=\tilde{g}$ because they agree on the dense subset $X$. Restriction and coordinate extension are inverse linear isometries, henceThey also preserve products: the corresponding continuous functions agree on $X$ and hence everywhere.

For the universal extension property, let $f:X\to K$ be continuous and $K$ a compact Hausdorff space. Pullback defines a bounded linear operator $T:C(K)\to C_b(X)$ by $Tg=g\circ f$. Its dual map $T^{\prime }:E^{\prime }\to C(K{)}^{\prime }$ is weak-star continuous, since $(T^{\prime }p)(g)=p(Tg)$. Evaluation ${\delta }_K:K\to C(K{)}^{\prime }$ is a homeomorphism onto a compact, thus closed, subset: a compact Hausdorff space is completely regular, by the Urysohn lemma. Now $T^{\prime }{\delta }_x={\delta }_K(f(x))$, soTherefore $\overline{f}={\delta }_K^{-1}\circ T^{\prime }{|}_{\beta X}$ is the required continuous extension. Two continuous maps into the Hausdorff space $K$ agreeing on the dense subset $X$ agree everywhere, proving uniqueness.

If $X$ is open in $\beta X$, then it is a locally compact space: around any $x$, regularity of the compact Hausdorff space $\beta X$ gives an open neighborhood whose closure is compact and lies in $X$. Conversely, suppose $X$ is locally compact. Choose an open neighborhood $U$ of $x$ in $X$ with compact closure $C\subseteq X$. There is an open $O\subseteq \beta X$ with $O\cap X=U$. Density of $X$ implies $O\subseteq {\overline{U}}^{\beta X}$, while compactness of $C$ makes it closed in $\beta X$, so ${\overline{U}}^{\beta X}\subseteq C$. Thus $x\in O\subseteq X$. This provesThe empty-space cases are immediate.

The final printed assertion, for an arbitrary subset of $\beta X$, is false. Take the infinite discrete space $X=\mathbb{N}$. It is not compact, so its compactification has a point $p\in \beta \mathbb{N}\setminus \mathbb{N}$. The singleton $\{p\}$ is closed, hence equals its own closure. It is not open, since every nonempty open subset of $\beta \mathbb{N}$ meets the dense copy of $\mathbb{N}$. This is a counterexample.

The valid result holds for subsets $D\subseteq X$, and more generally for open subsets of $\beta X$. For $D\subseteq X$, its indicator is bounded continuous because $X$ is discrete. Its extension $\widetilde{1_D}$ is still idempotent by density, so it takes only the values zero and one. The set $H=\{\widetilde{1_D}=1\}$ is clopen and has $H\cap X=D$. Density then gives $H={\overline{D}}^{\beta X}$. If $O$ is open in $\beta X$, the set $O\cap X$ is dense in $O$, so $\overline{O}=\overline{O\cap X}$ is clopen by the result just proved. Consequently $\beta X$ is extremally disconnected, which means the closure of every open set is open. This is the discrete Stone-Čech compactifications are extremally disconnected theorem, with the essential open-set qualification.

Write $C(K)={\bigcup }_{n\ge 1}{\mathcal{S}}_n$ with every ${\mathcal{S}}_n$ equicontinuous. Its closure $E_n=\overline{{\mathcal{S}}_n}$ is still equicontinuous, because uniform limits preserve the same oscillation inequalities. The Baire category theorem applied to the Banach space $C(K)$ shows that some $E_n$ has nonempty interior. Hence there are $f_0\in C(K)$ and $r>0$ such that the open ball $B(f_0,r)$ lies in $E_n$.

Fix $x\in K$. Equicontinuity of $E_n$ gives a neighborhood $U$ of $x$ on which every member of $E_n$ changes by less than $r/4$. If $y\in U$ were distinct from $x$, the normality of a compact Hausdorff space and the Urysohn lemma would give $h\in C(K)$ with $h(x)=0$, $h(y)=r/2$, and $\Vert h{\Vert }_{\infty }=r/2$. Both $f_0$ and $f_0+h$ belong to $E_n$, yeta contradiction. Thus $U=\{x\}$, so every point is isolated and $K$ is a discrete space. A compact discrete space is finite, since its cover by singleton open sets has a finite subcover.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.

The one-point compactification $D\cup \{\infty \}$ of an uncountable discrete space is a hereditarily normal space but not a perfectly normal space. A subspace missing $\infty$ is discrete; one containing it is closed, hence compact Hausdorff and normal. However, every neighborhood of $\infty$ misses only finitely many points of $D$. A countable intersection misses at most countably many points, so the closed singleton $\{\infty \}$ is not a G-delta set.