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The Tychonoff theorem states that an arbitrary product of compact spaces, with the product topology, is compact; no Hausdorff hypothesis is necessary. We prove it using Zorn's lemma and the axiom of choice.

First, every proper filter on a set extends to an ultrafilter. Order its proper extensions by inclusion. The union of a chain is again a proper filter: any finite collection of its members occurs in one member of the chain, and the empty set occurs in none. Zorn's lemma gives a maximal proper filter $\mathcal{U}$. If $A\notin \mathcal{U}$, adjoining $A$ must destroy properness, so some $F\in \mathcal{U}$ has $F\cap A=\varnothing$, and hence $X\setminus A\in \mathcal{U}$. This proves the decision property of an ultrafilter.

Next prove the ultrafilter characterization of compactness. In a compact space the closed sets $\overline{F}$, for $F\in \mathcal{U}$, have the finite intersection property, so their intersection contains a point $x$. Every open neighbourhood $O$ of $x$ belongs to $\mathcal{U}$: otherwise its closed complement belongs to $\mathcal{U}$ and contains $x$, a contradiction. Thus the ultrafilter converges to $x$. Conversely, if an open cover has no finite subcover, the closed complements have the finite intersection property and generate a proper filter. Extend it to an ultrafilter. A limit point lies in some member $O$ of the cover, so both $O$ and its complement would belong to the ultrafilter. This is impossible. These arguments also prove the closed-set form of compactness directly by taking complements.

Now let $X={\prod }_{j\in J}{X}_j$. If a factor is empty, the product is empty and compact. Otherwise the axiom of choice makes $X$ nonempty. For an ultrafilter $\mathcal{U}$ on $X$, its coordinate imageis an ultrafilter. Compactness of $X_j$ supplies at least one limit $x_j$; use the axiom of choice to choose one for every $j$. A basic open neighbourhood of $x=(x_j{)}_j$ restricts finitely many coordinates. Each corresponding inverse image belongs to $\mathcal{U}$, so their finite intersection also does. Hence $\mathcal{U}$ converges to $x$ in the product topology. The ultrafilter criterion proves the theorem. The explicit choices and the maximal-filter argument are exactly the places where choice enters.

The Banach-Alaoglu theorem states that the closed unit ball of the continuous dual space $V^{\ast }$ of a real or complex normed vector space $V$ is compact in the weak-star topology $\sigma (V^{\ast },V)$. Completeness of $V$ is not required.

For each $v\in V$, let $D_v=\{z\in \mathbb{K}:|z|\le \Vert v\Vert \}$, a compact scalar disk or interval. By the Tychonoff theorem, $P={\prod }_{v\in V}{D}_v$ is compact. Inside $P$, impose the equationsEach equation defines a closed subset, since it concerns continuous coordinate projections into a Hausdorff scalar field. Their intersection $K$ is therefore compact. A point in $K$ defines a linear functional $\ell (v)=z_v$, and the coordinate bounds give $|\ell (v)|\le \Vert v\Vert$. Conversely, every $\ell \in V^{\ast }$ with $\Vert \ell \Vert \le 1$ gives such a point. Thus $K$ is exactly the image of the dual unit ball under $\ell \mapsto (\ell (v){)}_{v\in V}$. The induced product topology is precisely pointwise convergence on $V$, which is the weak-star topology. This proves

For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate continuity and linearity make both maps bounded linear operators. For each fixed $y$,The Uniform boundedness principle therefore gives $C<\infty$ such that $\Vert L_x\Vert \le C$ whenever $\Vert x\Vert \le 1$. Scaling yieldsFor completeness, the relevant Uniform boundedness principle follows from the Baire category theorem as follows. For a pointwise bounded family $\mathcal{T}$ of bounded linear operators on a Banach space, the closed sets $E_m=\{y:{\sup }_{T\in \mathcal{T}}\Vert Ty\Vert \le m\}$ cover the space. One $E_m$ contains an open ball $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\Vert h\Vert <r$, gives ${\sup }_T\Vert Th\Vert \le 2m$. Applying this to $h=(r/2)u$, $\Vert u\Vert \le 1$, bounds every operator norm by $4m/r$.

Define the new norm byIt is a norm because $L_{x}e=x$, so $L_x=0$ implies $x=0$. The triangle inequality and scalar homogeneity follow from those of the operator norm. Moreover,so it is equivalent to the original norm and remains complete. By associativity, $L_{xy}=L_x{L}_y$, and the submultiplicativity of the operator norm provesThis is renorming a separately continuous Banach algebra. It even gives $\Vert e{\Vert }_{\ast }=1$. If the algebra is the zero space, the conclusion is immediate separately.

The sum of the two metrics is nonnegative and symmetric. It vanishes exactly when both coordinate distances vanish, hence exactly when the two points agree. Adding the two component triangle inequalities gives the triangle inequality on the Cartesian product, so the displayed function is a metric. Moreover,so the projection map is $1$-Lipschitz and therefore continuous.

To prove sequential compactness directly from compactness, take a sequence $(x_n)$ in $X$. If every point $x$ had a neighborhood containing sequence terms at only finitely many indices, these neighborhoods would form an open cover with a finite subcover, implying that only finitely many sequence terms exist, a contradiction. Thus some $x\in X$ has infinitely many indices in every neighborhood. Choose increasing indices $n_k$ with $d_X(x_{n_k},x)<1/k$; then $x_{n_k}\to x\in X$.

Now let $F\subset X\times Y$ be closed and suppose $y_n\in \pi (F)$ converges to $y$. Choose $x_n$ such that $(x_n,y_n)\in F$. A convergent subsequence $x_{n_k}\to x$ exists by the preceding argument. In the product metric, $(x_{n_k},y_{n_k})\to (x,y)$, and the fact that $F$ is closed gives $(x,y)\in F$. Thus $y\in \pi (F)$, proving that the projection with a compact factor is closed. The sequential criterion for a closed set in a metric space follows by choosing points at distance less than $1/k$ from any point of its closure.

For a counterexample without a compact factor, take $X=Y=\mathbb{R}$ with their usual metrics andThe continuous product function makes $F$ closed, but $\pi (F)=\mathbb{R}\setminus \{0\}$ is not closed. Thus compactness of the discarded factor is essential to the general conclusion.

For integral singular cochains $a\in C^p(X)$ and $b\in C^q(X)$, define the cup product on an ordered simplex byThe same construction works over any commutative coefficient ring. Expanding the coboundary as the alternating sum of faces gives the Leibniz identityIndeed faces deleting vertices before the split occur in the first sum, faces after the split occur in the second, and the two terms at the common vertex cancel with opposite signs. Thus the product of cocycles is a cocycle. Replacing $a$ by $a+\delta c$ changes the product with a cocycle $b$ by $\delta (c⌣b)$; replacing $b$ by $b+\delta d$ changes the product with a cocycle $a$ by $(-1{)}^p\delta (a⌣d)$. Hence the product is independent of representatives and defines a well-defined product on cohomology classes.

With ${\mathbb{F}}_2={\mathbb{Z}}_2$ coefficients, the real-projective-space cohomology ring isFor $n>m\ge 1$, the pullback of the degree-one generator is either zero or $t$. The second choice would send the zero class $t^{m+1}$ to the nonzero class $t^{m+1}\in H^{m+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)$, contradicting the naturality of the cup product. Thus the generator pulls back to zero, and so do all its positive powers. For $m=0$ the target's reduced cohomology is already zero. Therefore every such map induces zero on reduced mod-two cohomology.

Suppose a continuous $f:S^n\to {\mathbb{R}}^n$ had $f(x)\ne f(-x)$ everywhere. Thenis a continuous odd map to $S^{n-1}$ and descends to $\bar{h}:{\mathbb{RP}}^n\to {\mathbb{RP}}^{n-1}$. For $n\ge 2$, a path from $x$ to $-x$ projects to a loop detected by the degree-one generator of the double cover. Its image lifts to a path from $h(x)$ to $-h(x)$, so is detected by the target's double cover as well. Thus ${\bar{h}}^{\ast }$ is nonzero in degree one, contradicting the preceding calculation. For $n=1$, an odd continuous map from the connected circle to $S^0$ is already impossible. The case $n=0$ is immediate. HenceThis proves the Borsuk-Ulam theorem rather than assuming it.

For a closed cover $A_1,\dots ,A_{n+1}$, assume no member contains an antipodal pair. Apply the just-proved theorem to $F(x)=(d(x,A_1),\dots ,d(x,A_n))$, using any metric on the sphere. Empty sets can be assigned a constant positive coordinate. It gives equal distances at $x,-x$. If either point belongs to an $A_i$ with $i\le n$, zero distance and closedness place both points in it, a contradiction. Hence neither lies in any of the first $n$ sets, and both must lie in $A_{n+1}$, another contradiction. This proves the Lusternik-Schnirelmann-Borsuk theorem for closed sets.

Four closed sets on $S^2$ need not contain an antipodal pair. Take the four unit vertices $v_i$ of a regular tetrahedron centered at zero and define $C_i=\{x:x\cdot v_i\ge x\cdot v_j\text{ for every }j\}$. These closed sets cover the sphere by maximizing the four scalar products. If $x$ and $-x$ belonged to $C_i$, both inequalities would force all four scalar products to be equal. Since $\sum v_i=0$, they would all be zero, and because the vertices span ${\mathbb{R}}^3$, $x$ would be zero. This is impossible on the sphere. Thus these four sets give the antipodal-free closed cover from simplex Voronoi cells counterexample.

Write $p=1+\epsilon$ and assume $\lambda >0$. The large-deviation speed is $a_N=N^p$, and the answer isHere and below a large deviation principle at speed $a_N$ means the open-set lower bound $\mathrm{liminf}{a}_N^{-1}\log \mathbb{P}(X_N\in G)\ge -{\inf }_{G}I$ and the closed-set upper bound $\mathrm{limsup}{a}_N^{-1}\log \mathbb{P}(X_N\in F)\le -{\inf }_{F}I$. The displayed rate function is nontrivial and is a good rate function.

For $0<u<v$, the exponential distribution givesShrinking an interval about any $x>0$ supplies the local lower bound $-\lambda x$; neighborhoods of zero have probability tending to one. Every open set intersecting the nonnegative half-line contains one of these neighborhoods, giving the open-set bound. If a closed set $F$ contains zero, its upper bound is just $\mathbb{P}(\cdot )\le 1$. If it meets $[0,\infty )$ but excludes zero, closedness gives $b=\inf (F\cap [0,\infty ))>0$, and $\mathbb{P}(A/a_N\in F)\le e^{-\lambda a_{N}b}$. A closed set disjoint from the support has probability zero. This proves the full large deviation principle, including the boundary at zero.

The fixed exponential tail explains the speed: an excursion of size $x{N}^{1+\epsilon }$ costs $e^{-\lambda x{N}^{1+\epsilon }}$. The full Gärtner–Ellis theorem is unnecessary here; the limiting cumulant-generating function is flat on $\theta <\lambda$ and infinite at $\theta \ge \lambda$, so it fails the theorem's essential smoothness hypothesis.

The printed mean is $1/(\lambda i)$, so $T_i$ has exponential distribution of rate $i\lambda$. Let $S_k={\sum }_{i=1}^k{T}_i$, with fixed $k$. ThenHere is a direct proof, which also identifies the exceptional role of the slowest exponential clock. For $0<\theta <\lambda$, independence and the moment-generating function giveLetting $\theta \uparrow \lambda$ proves the tail upper bound $-\lambda x$. For a local lower bound at $x>0$, fix $0<\eta <x$ and $M>0$ with $\mathbb{P}({\sum }_{i=2}^k{T}_i\le M)>0$; for $k=1$ this probability is one. For large $N$, the eventputs $S_k/N$ in $(x-\eta ,x+\eta )$. Its logarithmic probability divided by $N$ tends to $-\lambda (x-\eta /2)$. Shrinking $\eta$ gives the local lower bound $-\lambda x$. At zero, $S_k/N\to 0$ almost surely and neighborhoods have probability tending to one.

For any closed set excluding zero, its nonnegative part is bounded away from zero; the tail upper bound applies at its infimum. Sets outside the support have probability zero. Together with the local lower bounds this proves the full large deviation principle, with a good rate function. Equivalently, the product large-deviation principle for the fixed vector $(T_1/N,\dots ,T_k/N)$ and the contraction principle for large deviations minimize ${\sum }_{i}i\lambda u_i$ subject to $u_i\ge 0$ and ${\sum }_i{u}_i=x$; all the excursion is assigned to $u_1$.

A linear operator between Banach spaces is a compact operator when it maps the unit ball into a relatively compact set, equivalently when every bounded sequence of inputs has an image subsequence converging in norm. Such an operator is bounded. For a bounded linear operator on a complex Banach space, the resolvent set consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the spectrum. The point spectrum consists of those $\lambda$ for which $T-\lambda I$ has a nonzero kernel, namely the eigenvalues.

On a Hilbert space, $T$ is self-adjoint if $T=T^{\ast }$, or equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all vectors. The given prescription extends uniquely toIt is bounded with norm one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_{m}x={\sum }_{j\le m}{x}_j{e}_j/j$. Then $\Vert T-T_m\Vert =1/(m+1)\to 0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence $T$ is compact and self-adjoint.

Each $1/j$ is an eigenvalue with eigenspace $\mathbb{C}{e}_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda )x_j=0$, so there are no other eigenvalues; in particular zero is not an eigenvalue. But $\Vert T{e}_j\Vert =1/j\to 0$, which contradicts a bounded inverse at zero. Thus zero is in the spectrum.

For $\lambda \notin \{0,1,1/2,1/3,\dots \}$, its distance from this closed set is positive. The diagonal formulathen defines a bounded inverse. This proves the complete description

The closed-set form of the Baire category theorem says: if a nonempty complete metric space is a countable union of closed sets, at least one of those sets has nonempty interior. Equivalently, a complete metric space cannot be a countable union of closed sets with empty interiors. In the application the space is $\mathbb{R}$ with its usual complete metric.

Assume the diagonal $\Delta \subset X\times X$ is closed. For continuous $f,g:Y\to X$, the map $h:Y\to X\times X$, $h(y)=(f(y),g(y))$, is continuous: the inverse image of a basic rectangle $U\times V$ is $f^{-1}(U)\cap g^{-1}(V)$. Thereforeis closed as the inverse image of a closed set under a continuous map. This proves (ii) implies (iii).

The closure $\mathrm{cl}(A)$ is the intersection of all closed subsets containing $A$; equivalently it consists of the points whose every open neighbourhood meets $A$.

Suppose $f$ is continuous and $x\in \mathrm{cl}(A)$. For any open neighbourhood $V$ of $f(x)$, continuity makes $f^{-1}(V)$ an open neighbourhood of $x$, so it meets $A$. Hence $V$ meets $f(A)$, showing $f(x)\in \mathrm{cl}(f(A))$. This proves $f(\mathrm{cl}(A))\subseteq \mathrm{cl}(f(A))$.

Conversely suppose this inclusion holds for every subset $A$. For a closed subset $C\subseteq Y$, let $A=f^{-1}(C)$. Then $f(A)\subseteq C$ and therefore $\mathrm{cl}(f(A))\subseteq C$. By the assumed inclusion, $f(\mathrm{cl}(A))\subseteq C$, so $\mathrm{cl}(A)\subseteq f^{-1}(C)=A$. Since $A\subseteq \mathrm{cl}(A)$ always, $A$ is closed. Thus every closed set has closed preimage, equivalently every open set has open preimage. This is continuity, proving the claimed equivalence.

We prove the Borel determinacy theorem by reducing each Borel game to a clopen game while preserving winning strategies. The construction permits arbitrary set-sized alphabets; no enumeration of the move set is used. Assume $A\ne \varnothing$. For an empty alphabet the game ends at its initial position, with its assigned terminal outcome, and is trivially determined.

Allow intermediate game trees $T$ to have terminal positions with assigned winners. Write $[T]$ for the space of infinite branches, with its prefix-cylinder topology; the payoff $B\subseteq [T]$ governs infinite plays, while the assigned winners govern finite plays. Question 4's attractor proof still gives open and clopen determinacy: include I-winning terminals among the initial targets and make the opposing terminals absorbing losses.

A covering of an infinite game consists of a new tree $\tilde{T}$, a prefix-preserving map $\pi :\tilde{T}\to T$ preserving lengths, and a strategy map $\Phi$ for each player. Old terminal outcomes must be inherited. The strategy map is local in depth: choices before depth $n$ depend only on the given strategy before depth $n$. Its essential lifting requirement is that each original play $x$ following $\Phi (\sigma )$ has a lift $\tilde{x}$ following $\sigma$ whose projection is $x$, or whose projection is an initial segment of $x$ and which ends with a loss for the player using $\sigma$. Infinite projections are taken by union of prefixes. A $k$-covering is identical to the original tree, terminal assignments and strategy choices through depth $k$.

These conditions transfer winning strategies for every payoff. Indeed, if $\sigma$ wins the covered game with payoff ${\pi }^{-1}(B)$, a lift of a play following $\Phi (\sigma )$ cannot be an early loss for its player. The lift therefore projects to the whole original play. Infinite membership in $B$ agrees under projection, and an original terminal loss would also be a loss in the game covering. Hence the original play is won. Call a game covering an unravelling of a game payoff when ${\pi }^{-1}(B)$ is clopen in $[\tilde{T}]$.

We first construct a $k$-covering unravelling any closed $C\subseteq [T]$. Choose an even $m\ge k$ and leave the first $m$ levels unchanged. At a nonterminal position $p$ of length $m$, I's next move will carry auxiliary information. For each legal ordinary move $a$, let $Z$ be the set of first nonterminal positions strictly beyond $pa$ whose infinite continuations miss $C$: $q\in Z$ means $q$ extends $pa$, $[T_q]\cap C=\varnothing$, and no intervening strict extension of $pa$ has this property. These positions are pairwise incomparable. I plays $(a,X)$, where $X\subseteq Z$.

If $pa$ was already terminal, inherit its outcome. Otherwise II has two types of response. On accepting, II plays $(\mathrm{accept},b)$ with $b$ an ordinary legal move, and subsequent moves are ordinary. When a position $q\in Z$ is reached, end the covered game with a win for I if $q\in X$ and for II if $q\notin X$. Other old terminals retain their winners. On challenging, II plays $(\mathrm{challenge},q,b)$ with $q\in X$ and $pab$ an initial segment of $q$; subsequent play is forced along $q$ until reaching it and then proceeds ordinarily in $T_q$. Original terminal outcomes remain in force. Forget the auxiliary data to define $\pi$.

For an infinite accepted play, no exit in $Z$ was reached. Its projection belongs to $C$: if it did not, closedness would give a prefix cylinder disjoint from $C$, and the first such nonterminal extension strictly beyond $pa$ would have ended the play. An infinite challenged play projects through an exit $q$ and therefore misses $C$. Thus, among infinite plays, the inverse image of $C$ is exactly the acceptance branch of this finite protocol. It is clopen, decided at depth $m+2$.

Here are both strategy-transfer maps; these are necessary because continuous projection by itself does not preserve determinacy. Given a strategy $\sigma$ for I in the covered tree, follow its ordinary moves while simulating acceptance of its announcement $X$. If an exit $q\notin X$ is reached, continue arbitrarily: that play has an early covered loss for I as a lift. If $q\in X$ is reached, switch to the continuation prescribed by $\sigma$ under the challenge of $q$. This is consistent with the already observed prefix, since the challenged game forced every intermediate move along $q$. If no exit is reached, acceptance itself supplies the full lift. Old terminal plays lift in the same way. This defines $\Phi (\sigma )$ and verifies the lifting requirement for I.

For a covered strategy $\tau$ for II, at the projected position $pa$ defineAgainst announcement $Y$, the strategy must accept: a challenge would have to name an element of $Y$, contradicting its definition. Simulate this acceptance, including its ordinary response $b$. If an exit $q\in Y$ is reached, it is an early covered loss for II, so continue arbitrarily. If an exit $q\notin Y$ is reached, choose an announcement $X_q$ for which $\tau$ challenges $q$, and switch to that simulation. The challenged game's forced prefix agrees with the observed prefix, including the earlier ordinary response $b$, so it supplies a full lift following $\tau$. If no exit is reached, the original acceptance simulation gives the full lift. This defines $\Phi (\tau )$ and verifies lifting for II. Choices of $X_q$ use the axiom of choice on a set of possibilities.

Both maps are local in depth. The sets $Y$ and the chosen $X_q$ use only II's response function at depth $m+1$; later choices use only strategy values at the already reached depth. The switches do not query choices at later levels. On irrelevant histories assign fixed legal moves. Thus this is a genuine $k$-covering. Complementing $C$ changes neither the game covering nor its lifting property, and complements of clopen sets are clopen. This proves the base construction for both closed and open payoff sets.

We next give the limiting construction needed for the induction. Coverings compose by composing their position and strategy maps; lifting successively verifies the same lifting condition. Supposeis a sequence of game coverings whose $i$th map leaves depth $k+i$ unchanged. For each finite depth $d$, the trees and terminal assignments through that depth eventually stop changing. Define $T_{\infty }$ using these stabilized levels. Its map to $T_i$ on a position of length $d$ is the composite from any sufficiently late stage; stabilization makes the value independent of the stage. Define strategy maps in the same way, using their locality in depth. These maps respect the tree, terminal outcomes and composition, and leave depth $k$ unchanged.

To check lifting, start with a play of $T_i$ consistent with the projected strategy. Lift the entire play successively through the later game coverings, using the corresponding projected strategy at each stage. If all these lifts are infinite, each projects to the preceding one and each finite prefix eventually stabilizes, giving the desired infinite lift in $T_{\infty }$. If a premature own-player loss occurs, later lifts have nonincreasing finite lengths, and those lengths and their prefixes eventually stabilize. This gives a terminal own-player loss in $T_{\infty }$, which is exactly the allowed exceptional lift. The same reasoning handles original finite terminal plays. Thus $T_{\infty }$ covers every $T_i$. This argument uses level stabilization, not compactness of an infinitely branching tree.

Now induct on the countable Borel hierarchy rank, with the strengthened assertion: for every game tree on any set alphabet and every $k$, a payoff of that rank has a $k$-covering unravelling it. The base case of open or closed sets was proved above. Complements preserve the assertion using the same game covering. For the union step, write $B={\bigcup }_{n<\omega }{B}_n$ with constituent sets of smaller rank. Starting at $T_0=T$, successively choose a $(k+n)$-covering $T_{n+1}\to T_n$ that unravels the inverse image of $B_n$ under the composite map to $T$. This is legitimate because continuous inverse images preserve Borel hierarchy rank, and the induction assertion ranges over all trees and alphabets, including the enlarged ones produced by earlier game coverings.

In the stabilized inverse limit, each pulled-back $B_n$ is clopen, because the later projection to the stage where it was unravelled is continuous. Their union is therefore open. Apply the already proved open-set unravelling construction once more, with preservation depth $k$, and compose with the limit game covering. The result unravels $B$ itself. This proves the strengthened assertion at every successor or limit countable Borel hierarchy rank. Every Borel set arises at some such rank: countable unions take the supremum of countably many countable ranks and then one further stage, and complements do not leave the generated hierarchy.

Apply the assertion to $T=A^{<\omega }$ and $k=0$. The covered payoff is clopen and hence determined by Question 4, including its assigned finite terminal outcomes. Transfer its winning strategy through the game covering. We obtainEvery auxiliary alphabet and strategy collection used above is a set. The argument uses power sets and replacement to build those collections and their iterated game coverings, and the axiom of choice for strategy selections; it does not require $A$ to be countable.

The closed-set version of the Baire category theorem says that if a nonempty complete metric space is a countable union of closed sets, at least one of those closed sets has nonempty interior. Equivalently, a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

Take $f(0)=0$ and $f(x)=1/x$ for $x\ne 0$. Its graph is $\{(x,y):xy=1\}\cup \{(0,0)\}$, a union of two closed sets, since multiplication is continuous. But $f(1/n)=n$ does not tend to $f(0)$, so the function is not continuous.

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.

For each integer $n\ge 0$, put $Z_n=\{z:f^{(n)}(z)=0\}$. Every derivative $f^{(n)}$ is an entire function and cannot be identically zero: otherwise repeated integration would make $f$ a polynomial. By the identity theorem for holomorphic functions, its zero set has empty interior, and continuity makes $Z_n$ a closed set. Thus $Z_n$ is nowhere dense.

The complex plane with its usual metric is a complete metric space, so the Baire category theorem gives $z_0\notin {\bigcup }_{n\ge 0}{Z}_n$. The Taylor series there has coefficients

The closed-set version of the Baire category theorem says that if a nonempty complete metric space $X$ is a countable union $X={\bigcup }_{n=1}^{\infty }{F}_n$ of closed sets, then at least one $F_n$ has nonempty interior. Equivalently, a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal space admit a continuous $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. Using the usual convention, normal spaces are also $T_1$.

The Tietze extension theorem states that a continuous real-valued function on a closed subset of a normal space extends continuously to the whole space. If its range lies in a closed bounded interval, the extension may be chosen in that same interval.

Prove first the interval version for $f:S\to [-1,1]$. If a continuous residual $r$ on $S$ obeys $|r|\le M$, the sets $\{r\le -M/3\}$ and $\{r\ge M/3\}$ are disjoint closed subsets of $X$. Urysohn's lemma gives a continuous $h:X\to [-M/3,M/3]$ taking the endpoint values on those two sets. On each of the three ranges of $r$, it follows that $|r-h|\le 2M/3$ on $S$. Repeating with $M_j=(2/3{)}^j$ produces $h_j$ with $\Vert h_j{\Vert }_{\infty }\le M_j/3$ and a residual bounded by $M_{j+1}$. The series $\tilde{f}={\sum }_{j\ge 0}{h}_j$ converges uniformly, has absolute value at most $\sum M_j/3=1$, is continuous and restricts to $f$ as the residual tends to zero. Rescaling proves every closed bounded interval version.

For unbounded $f$, first extend $f/(1+|f|)$ to $g:X\to [-1,1]$. The closed set $B=g^{-1}(\{-1,1\})$ is disjoint from $S$. Urysohn's lemma gives $\psi :X\to [0,1]$ with $\psi {|}_S=1$ and $\psi {|}_B=0$. Then $h=\psi g$ takes values strictly between minus one and one everywhere and equals $f/(1+|f|)$ on $S$. Thus $h/(1-|h|)$ is a continuous real-valued extension of $f$, proving the full theorem.

Statement 1 is true: metric spaces are normal, and the bounded interval form applies. Statement 2 is true: compact Hausdorff spaces are normal, and the unrestricted real-valued form applies. Statement 3 is false. For a counterexample, let $X=\{a,b,c\}$ have open sets $\varnothing ,\{a\},\{a,b\},\{a,c\},X$. The subset $S=\{b,c\}$ is closed and discrete, so $f(b)=-1$, $f(c)=1$ is continuous on it. Every open neighborhood of $b$ or $c$ in $X$ contains $a$. A continuous extension would pull back disjoint neighborhoods of minus one and one to disjoint open sets containing $b$ and $c$, yet both would contain $a$. This is impossible.

False without a separation assumption. Give $X=\{0,1\}$ the indiscrete topology $\{\varnothing ,X\}$, and take $Y=\{0\}$. Both spaces are finite and hence compact, but $Y$ is not a closed set because $X\setminus Y=\{1\}$ is not open. In a Hausdorff space, a compact subset would be closed; that hypothesis is absent here.

In $\mathbb{R}$, takeThese are nonempty descending sets of diameter $1/n$, with empty intersection. They are not closed sets, so there is no contradiction with the Cantor intersection theorem for the complete metric space $\mathbb{R}$.

A metric space is complete if every Cauchy sequence in it converges to a point of that space. The stated shrinking-set characterization is the Cantor intersection theorem together with its converse.

First assume completeness, and choose $x_n\in A_n$. If $m,n\ge N$, then both points belong to $A_N$, so $d(x_m,x_n)\le \mathrm{diam}(A_N)$. The vanishing diameters imply that $(x_n)$ is a Cauchy sequence, hence it converges to some $x\in X$. For every fixed $N$, its tail belongs to the closed set $A_N$, so $x\in A_N$. Thus $x\in {\bigcap }_N{A}_N$. In fact the intersection contains only one point, since any two of its points have distance at most every diameter.

Conversely assume the stated intersection property and let $(x_n)$ be a Cauchy sequence. PutThese closures are nonempty, closed and nested. Taking a closure does not increase the diameter: approximate two points of the closure by points of the set and use continuity of distance. The Cauchy sequence therefore makes $\mathrm{diam}(A_N)\to 0$. Choose $x$ in the intersection. For $n\ge N$, both $x_n$ and $x$ belong to $A_N$, and $d(x_n,x)\le \mathrm{diam}(A_N)\to 0$. Every Cauchy sequence converges in $X$, proving completeness.