Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2007 (C1.2, DMAC 2461, DMPC 2461)

Use these standard ultrafilter facts: a space is compact if and only if every ultrafilter on it converges; the image ultrafilter under a map is an ultrafilter; and a convergent ultrafilter contains every neighborhood of its limit.

Let $X={\prod }_{\lambda \in \Lambda }{X}_{\lambda }$ with all factors compact. An empty factor makes $X$ empty and hence compact. Otherwise, given an ultrafilter $\mathcal{U}$ on $X$, push it forward by each product projection to obtain an ultrafilter ${\mathcal{U}}_{\lambda }$ on $X_{\lambda }$. Choose a limit $x_{\lambda }$ of each, and put $x=(x_{\lambda }{)}_{\lambda }$. A basic product neighborhood of $x$ is an intersection of finitely many inverse images of coordinate neighborhoods. Each inverse image belongs to $\mathcal{U}$, so their finite intersection does too. Thus $\mathcal{U}$ converges to $x$, proving compactness of the product. Uniqueness of coordinate limits is not needed, so the factors need not be Hausdorff.

For local compactness, suppose only the factors indexed by a finite set $F$ may be noncompact. At $x\in X$ choose, for each $\lambda \in F$, a compact neighborhood $K_{\lambda }$ of $x_{\lambda }$, with an open neighborhood $U_{\lambda }\subseteq K_{\lambda }$. For other indices put $K_{\lambda }=X_{\lambda }$. The compact product ${\prod }_{\lambda }{K}_{\lambda }$ contains the open product neighborhood restricting just the finitely many coordinates in $F$ to $U_{\lambda }$. Hence every point has a compact neighborhood and the product is locally compact.

The unrestricted assertion is false. Each factor of ${\mathbb{R}}^{\mathbb{N}}$ is locally compact, but its product is not. Any basic neighborhood restricts only finitely many coordinates. Choose a free coordinate $j$; its projection of that neighborhood is all of $\mathbb{R}$. A compact neighborhood containing it would consequently have compact image equal to $\mathbb{R}$, impossible. This shows directly that no point has a compact neighborhood.

For $x$ in a compact Hausdorff space $K$, its quasi-component $Q$ is the intersection of all clopen sets containing $x$. It is closed and therefore compact. Suppose $Q=A\bigsqcup B$ is a separation of a topological space into two nonempty relatively closed sets, with $x\in A$. The sets $A,B$ are compact and hence closed in $K$. Normality of a compact Hausdorff space gives disjoint open neighborhoods $U,V$ of $A,B$.

The compact set $K\setminus (U\cup V)$ is disjoint from $Q$. For each point $y$ of that set choose a clopen $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus (U\cup V)$, so their corresponding intersection $C$ is clopen, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is clopen in $C$, hence clopen in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus $Q$ is connected. Any connected set containing $x$ lies in every clopen neighborhood of $x$, so the connected component of $x$ equals $Q$.

For the second assertion, $D$ is closed in the ambient Hausdorff space $X$, and is itself compact Hausdorff. Let $C$ be a connected component of $D$ disjoint from ${\partial }_{X}D$. Choose $p\in C$. By the first assertion $C$ is the quasi-component of $p$ in $D$. For each $y\in {\partial }_{X}D$ choose a clopen subset of $D$ containing $p$ but not $y$. Compactness of ${\partial }_{X}D\subseteq D$ gives a finite intersection $H$ of these sets which contains $C$ and avoids the whole boundary. Thus $H\subseteq {\mathrm{int}}_{X}D$. It is closed in $X$, because it is closed in the closed set $D$, and is open in $X$, because it is relatively open in $D$ and contained in its interior. It is nonempty and proper, contradicting connectedness of $X$. Hence every connected component of $D$ meets its boundary.

Finally suppose a closed subset $A$ of the square has the point-separating clopen property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial line segment: it contains an interior point of the square and a small ball about that point. If such a segment lay in $A$, a clopen subset of $A$ separating its endpoints would induce a separation of a topological space of the connected segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is nowhere dense in the square, and also nowhere dense in ${\mathbb{R}}^2$.

Write $\alpha X=X\cup \{{\infty }_X\}$. Its open sets are the open subsets of $X$ together with sets $\alpha X\setminus K$, where $K$ is a compact subset of $X$. Compact subsets of the Hausdorff space $X$ are closed. This fact, and finite unions of compact sets, verify the topology axioms, including intersections of an old open set with a neighborhood of infinity. The inclusion $h_1:X\to \alpha X$ is a topological embedding with open image.

An open cover of $\alpha X$ has a member containing infinity; its complement is compact in $X$ and therefore is covered by finitely many other members. This proves compactness. Two points of $X$ can be separated in $X$. A point $x$ and infinity can be separated by an open neighborhood $U$ of $x$ contained in a compact neighborhood $K$, and $\alpha X\setminus K$. Thus $\alpha X$ is Hausdorff. Since $X$ is noncompact, every neighborhood of infinity meets $X$, so the embedded copy of $X$ is dense. This proves it is a Hausdorff compactification.

For a proper map, use the standard locally compact Hausdorff convention: it is continuous and inverse images of compact sets are compact. DefineAn open subset of $Y$ has open inverse image in $X$, hence in $\alpha X$. A basic neighborhood $\alpha Y\setminus K$ of infinity has inverse image $\alpha X\setminus f^{-1}(K)$, open because $f^{-1}(K)$ is compact. Hence $\alpha f$ is continuous and the diagram commutes. Any continuous map from a compact space to a Hausdorff space is proper: a compact subset of the target is closed, its inverse image is closed in the compact domain and therefore compact. This proves the required properness.

The last requested extension does not exist, even as a continuous map. Both sequences $h_1(0,n)$ and $h_1(1,n)$ tend to ${\infty }_{{\mathbb{R}}^2}$, since they eventually leave every compact subset of the plane. Their prescribed images would be the constant sequences $h_2(0)$ and $h_2(1)$. Continuity would force both distinct points to equal $g({\infty }_{{\mathbb{R}}^2})$, contradicting uniqueness of limits in the Hausdorff compactification of the line. The original product projection is not proper, as its inverse image of a singleton is an unbounded line.

If $\mathcal{B}$ is a countable topological basis of $X$, then $\{B\cap Y:B\in \mathcal{B}\}$ is a countable topological basis of the subspace $Y$. Indeed every subspace-open neighborhood is $Y\cap U$ for an open $U$ of $X$, and a suitable basic set lies between its point and $U$. This proves the first assertion.

Urysohn's metrisation theorem: every second-countable space that is also a T3 space is a metrizable space. Here regularity of a topological space allows a neighborhood of a point to be shrunk to an open neighborhood whose closure is still inside the original one.

We first prove the separation facts needed in a complete proof. Second countability implies the Lindelöf property: for each basic set contained in some member of an open cover choose one such member; these countably many members cover. In particular closed subspaces are Lindelöf. For disjoint closed sets $F,G$, regularity of a topological space and this property give open covers $F\subseteq {\bigcup }_n{U}_n$, $G\subseteq {\bigcup }_n{V}_n$ with ${\overline{U}}_n\cap G=\varnothing$ and ${\overline{V}}_n\cap F=\varnothing$. DefineThese sets are open and contain $F,G$ respectively. They are disjoint: if a point belonged to a term indexed by $n$ in the first union and by $m$ in the second, the term with the larger index would have removed the closure of the other covering set. Thus the space is a normal space.

For completeness, Urysohn's lemma supplies continuous separating functions in a normal topological space. Its construction is as follows. Given disjoint closed $F,G$, choose open sets $O_r$ for dyadic $r\in [0,1]$, with $F\subseteq O_0$, $O_1=X\setminus G$, and ${\overline{O}}_r\subseteq O_s$ whenever $r<s$. Start by shrinking a neighborhood of $F$ inside $X\setminus G$, and insert the successive dyadic sets by the closed-neighborhood shrinking in a normal space. DefineThen $f=0$ on $F$, $f=1$ on $G$, and $0\le f\le 1$. The identities $\{f<a\}={\bigcup }_{r<a}{O}_r$ and $\{f>a\}={\bigcup }_{a<r<1}(X\setminus {\overline{O}}_r)$ prove continuity for $0<a<1$, with the endpoint cases immediate.

Let $\{B_n\}$ be a countable topological basis. For each pair $(n,m)$ with ${\overline{B}}_n\subseteq B_m$, choose a continuous function $f_{nm}:X\to [0,1]$ equal to one on ${\overline{B}}_n$ and zero outside $B_m$. These form a countable family. Given $x\in O$ open, choose $B_m$ with $x\in B_m\subseteq O$, shrink a neighborhood of $x$ inside $B_m$, and then choose $B_n$ inside that smaller neighborhood. ThusConsequently these functions generate the original topology. They also separate distinct points: use the $T_1$ open set $X\setminus \{y\}$ when $x\ne y$. Enumerate them as $f_j$, padding by zero functions if necessary. The map $x\mapsto (f_j(x){)}_j$ is a topological embedding into the countable cube. Pulling back its metric, or directly usinggives a compatible metric. The fact that the functions distinguish distinct points proves positivity for distinct points, the triangle inequality is termwise, and the uniformly bounded tails show that this metric induces exactly the topology generated by the coordinates. Empty and singleton spaces are trivially metrizable spaces. This proves the theorem.

For the application, a separable metric space has a countable topological basis consisting of balls with rational positive radii about a countable dense set, and it is a T3 space. The given product theorems make a countable product of these spaces second countable and a T3 space. Urysohn's theorem makes it a metrizable space. Finally every second countable space is separable: choose a point from each nonempty basic open set; the resulting countable set meets every nonempty open set. Hence the product is separable and a metrizable space.

Here $Z{F}^{\ast }$ denotes ZF without foundation. Transfinite recursion on ordinals defines the cumulative hierarchy byThe associated class is $V=\{x:\exists \alpha \in \mathrm{Ord}(x\in V_{\alpha })\}$. Each level is a set, while $V$ is a definable class. Without Foundation it need not contain every ambient set; it consists of the sets built by this well-founded hierarchy.

Use transfinite induction. The inclusion at zero is immediate. If $V_{\beta }\subseteq V_{\beta +1}$, monotonicity of the power set givesIf $\alpha$ is a limit and $x\in V_{\alpha }$, choose $\beta <\alpha$ with $x\in V_{\beta }$. The inductive inclusion $V_{\beta }\subseteq \mathcal{P}(V_{\beta })$ means that $x\subseteq V_{\beta }\subseteq V_{\alpha }$. Thus $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. This proves $V_{\alpha }\subseteq V_{\alpha +1}$ for every ordinal. These successive inclusions and the union definition also give $V_{\beta }\subseteq V_{\alpha }$ whenever $\beta \le \alpha$.

The preceding inclusion is exactly $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$. Therefore every $x\in V_{\alpha }$ is a subset of $V_{\alpha }$. Equivalently,which says that $V_{\alpha }$ is a transitive set. This uses the power set definition and the proved inclusion, without any ambient use of Foundation.

We prove the stronger ordinal identityby transfinite induction. It holds at zero. At a successor stage an ordinal $\gamma$ belongs to $\mathcal{P}(V_{\beta })$ exactly when every $\delta <\gamma$ belongs to $V_{\beta }$. By induction this means every $\delta <\gamma$ is less than $\beta$, equivalently $\gamma \le \beta$. Thus the ordinals in $V_{\beta +1}$ are precisely those less than $\beta +1$. At a limit the identity follows by taking unions.

Every $\beta <\alpha$ consequently belongs to $V_{\alpha }$, so $\alpha \subseteq V_{\alpha }$ and $\alpha \in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. But $\alpha \notin V_{\alpha }$, since an ordinal is not less than itself. HenceThe only well-foundedness used is that of the ordinal order, available independently of Foundation for arbitrary ambient sets.

Every set is a subset of itself, so $V_{\alpha }\subseteq V_{\alpha }$. By the successor definition,Membership and inclusion are different statements here; this argument establishes membership of the whole level in its successor.

The class $V$ is transitive because all its levels are transitive. Let $a\in V$ be nonempty. There is a least ordinal $\alpha$ for which $a\cap V_{\alpha }\ne \varnothing$: existence follows by choosing an element of $a$, which lies in $V$, and then minimizing within some witnessing ordinal bound. The least ordinal is neither zero nor a limit, so write $\alpha =\beta +1$. Choose $x\in a\cap V_{\beta +1}$. Then $x\subseteq V_{\beta }$, while minimality gives $a\cap V_{\beta }=\varnothing$. Thus $a\cap x=\varnothing$. Also $x\in V$, proving Foundation inside $⟨V,\in ⟩$.

By the given hypothesis this same class satisfies every axiom of $Z{F}^{\ast }$, and hence it satisfies every axiom of $ZF$. Translate any finite formal $ZF$ proof by restricting each quantifier to the definable class $V$. Within $Z{F}^{\ast }$ the translated axioms have proofs, logical steps remain valid, and the translated contradiction is still a contradiction. Substituting those axiom proofs therefore converts the original derivation effectively into a finite derivation of contradiction from $Z{F}^{\ast }$. Equivalently,This is the syntactic relative-consistency construction; it does not assume that $V$ is a set or that it equals the whole ambient universe.

The transitive closure $\mathrm{TC}(a)$ is the least transitive set containing $a$ as a subset. Put $a_0=a$, $a_{n+1}=\bigcup a_n$ and defineRecursion and Replacement give the set of iterates, and Union gives its union. It contains $a=a_0$ as a subset. If $u\in v\in \mathrm{TC}(a)$, then $v\in a_n$ for some $n$, so $u\in a_{n+1}\subseteq \mathrm{TC}(a)$; hence it is transitive.

If $T$ is transitive with $a\subseteq T$, induction gives $a_n\subseteq T$ for every $n$, because $\bigcup T\subseteq T$. Therefore $\mathrm{TC}(a)\subseteq T$. This proves minimality and uniqueness. If the alternative convention asks for the least transitive set containing $a$ as an element, that set is $\mathrm{TC}(\{a\})$.

The membership induction principle is the schemaallowing fixed parameters in $\phi$.

Suppose the progressiveness hypothesis holds but $\neg \phi (a)$. The transitive set $T=\mathrm{TC}(\{a\})$ contains $a$. Separation gives the nonempty set $B=\{x\in T:\neg \phi (x)\}$. By Foundation choose $b\in B$ with $b\cap B=\varnothing$. Every $y\in b$ belongs to $T$ because it is a transitive set and cannot belong to $B$, so every such $y$ satisfies $\phi$. Progressiveness now gives $\phi (b)$, a contradiction. Thus the principle holds. The transitive closure is needed to turn a possibly proper class of counterexamples into a set to which Foundation applies.

An $\in$-isomorphism is a bijection $f:X\to Y$ such that $x\in x^{\prime }$ if and only if $f(x)\in f(x^{\prime })$ for all $x,x^{\prime }\in X$. When $Y$ is transitive, surjectivity gives the stronger recursive descriptionIndeed every member of $f(x)$ belongs to $Y$ because it is a transitive set and so has a preimage in $X$; the isomorphism property identifies exactly which preimages occur.

If $f,g$ are two such maps, possibly to different transitive sets, form $B=\{x\in X:f(x)\ne g(x)\}$. If $B$ were nonempty, Foundation would give $x\in B$ with $x\cap B=\varnothing$. All predecessors $u\in x\cap X$ then satisfy $f(u)=g(u)$, and the recursive descriptions force $f(x)=g(x)$, a contradiction. Therefore the map and its transitive range are unique.

The necessary and sufficient condition is that $⟨X,\in ⟩$ be well-founded and extensional as a relation on $X$. Well-foundedness says every nonempty $A\subseteq X$ has a member with no $\in$-predecessor in $A$. The extensional relation condition says distinct elements $x,x^{\prime }\in X$ have distinct predecessor sets $x\cap X$, $x^{\prime }\cap X$. Sufficiency is the Mostowski collapse theorem.

For necessity, an isomorphism transfers well-foundedness from its transitive range, where Foundation applies. For extensionality, if $x\cap X=x^{\prime }\cap X$, the recursive descriptions give $f(x)=f(x^{\prime })$, and injectivity gives $x=x^{\prime }$. Notice that ambient axiom of extensionality alone does not establish this internal predecessor condition; for example $X=\{\{0\},\{2\}\}$ can have distinct elements with no predecessors in $X$.

A cardinal number is an initial ordinal: an ordinal not in bijection with any smaller ordinal. In $ZFC$ every set has the cardinality of a unique such ordinal.

For an infinite cardinal $\kappa$, its cofinality is the least cardinality of an unbounded subset of the ordinal $\kappa$. Equivalently,A subset is unbounded when no ordinal below $\kappa$ bounds all its elements. A cofinal sequence can be taken increasing, and $\mathrm{cf}(\kappa )$ is an infinite regular cardinal.

König theorem for cardinal numbers says that for families of cardinals ${\kappa }_i<{\lambda }_i$ indexed by a set $I$,The sum is the cardinality of the disjoint union of sets $A_i$ of sizes ${\kappa }_i$, and the product is the cardinality of ${\prod }_i{B}_i$ for sets of sizes ${\lambda }_i$.

Consider any function $F:{⨆}_i{A}_i\to {\prod }_i{B}_i$. For each $i$, the set of $i$th coordinates of images of points from the $i$th block has cardinality at most ${\kappa }_i<{\lambda }_i$. Choose $b_i\in B_i$ outside it. Then $b=(b_i{)}_i$ is not the image of any point: a point from block $i$ has the wrong $i$th coordinate. Thus there is no surjection from the sum set onto the product set. Cardinal comparability in $ZFC$ makes the inequality strict in the indicated direction. For the empty index set the statement is $0<1$, also true. This is the diagonal argument underlying the inequality.

First derive the standard consequence of König theorem for cardinal numbers that every infinite cardinal $\rho$ satisfiesLet $\theta =\mathrm{cf}(\rho )$ and take cofinal ordinals ${\alpha }_i<\rho$, $i<\theta$. Put ${\rho }_i=\max (1,|{\alpha }_i|)<\rho$. Their sum is $\rho$: it is at least the size of the union of the cofinal ordinals and at most $\rho \cdot \theta =\rho$. Choose the next larger cardinal ${\rho }_i^{+}\le \rho$. König theorem for cardinal numbers givesThis also covers $\rho ={\aleph }_0$, using finite cardinals and their next larger finite cardinals.

Suppose $\lambda ={\kappa }^{\mu }$ with both $\kappa ,\mu$ infinite. Infinite cardinal multiplication givesIf $\mathrm{cf}(\lambda )\le \mu$, monotonicity of exponentiation would contradict the preceding strict inequality for $\rho =\lambda$. Thus $\mathrm{cf}(\lambda )>\mu \ge {\aleph }_0$. This direction does not require GCH.

Conversely assume GCH and $\mathrm{cf}(\lambda )>{\aleph }_0$. Every function $\omega \to \lambda$ has bounded range. For each infinite cardinal $\nu <\lambda$, GCH givesFinite bounded ranges are also covered by the bound ${\aleph }_0^{{\aleph }_0}\le {\aleph }_1\le \lambda$. Counting functions according to an ordinal bound on their range now gives ${\lambda }^{{\aleph }_0}\le \lambda \cdot \lambda =\lambda$. Constant functions give the reverse inequality. Thereforewhich is the required representation with $\kappa =\lambda$, $\mu ={\aleph }_0$. This proves both directions.

For a set $A$, its definable power set isOnly finitely many parameters occur in each definition. Satisfaction for the set structure is definable, so this specifies a uniformly definable class function assigning a set to every set $A$, with $\mathrm{Def}(A)\subseteq \mathcal{P}(A)$. Every subset definable over the structure with its parameters occurs. The full set $A$ itself occurs by the formula $x=x$. If $A$ is transitive, then $A\subseteq \mathrm{Def}(A)$, since any $p\in A$ is defined as a subset of $A$ by $x\in p$. In that case $\mathrm{Def}(A)$ is transitive as well: a member of one of its elements lies in $A$, hence in $\mathrm{Def}(A)$.

The constructible hierarchy and universe areThe displayed properties prove by induction that levels are transitive and increasing, so $L$ is transitive. It contains every ordinal: inductively $L_{\alpha }\cap \mathrm{Ord}=\alpha$, and the subset of $L_{\alpha }$ consisting of ordinals is definable there, giving $\alpha \in L_{\alpha +1}$. Ordinalhood is absolute here because the levels are transitive. Both the level function and membership in $L$ are definable in $ZF$. This is Gödel's constructible universe, which can contain fewer subsets than the full universe.

For each $x\in a$, let ${\beta }_x$ be the least ordinal with $x\in L_{{\beta }_x}$. It exists because $a\subseteq L$, and it is uniquely definable. Replacement therefore gives the set $\{{\beta }_x:x\in a\}$. Choose an ordinal $\alpha$ at least their supremum. Increasingness of the hierarchy givesIf $a$ is empty, $\alpha =0$ works. No assumption that $a$ itself belongs to $L$ is used; it is enough that its elements do.

Fix $a\in L$. In the ambient universe, Separation forms the setIt is a set of constructible sets, so the previous part gives an ordinal bound on all its members. Increase the bound to a level $L_{\beta }$ also containing $a$. Transitivity ensures $a\subseteq L_{\beta }$. ThenThe inclusion test is absolute for the transitive level $L_{\beta }$, since all members of $x$ and $a$ lie there. The right side is consequently a subset definable over $L_{\beta }$ with parameter $a$, soFor $x\in L$, transitivity of the constructible universe makes internal and ambient subsethood of $a$ agree. Hence $L$ satisfies that $S$ consists exactly of all its subsets of $a$. This is the Power set Axiom in $⟨L,\in ⟩$. It is not the assertion that every ambient subset of $a$ belongs to $L$.

Use the constructible universe theorem that $L$ satisfies $ZFC$ and GCH. In particular, internally it has a bijectionThis bijection belongs to $L$ and is also an actual bijection in the ambient universe, because its domain, range and graph are transitive-model sets with the same membership relation.

We must compare the ordinals rather than assume cardinality is absolute. Every ordinal below ${\omega }_1^L$ is countable in $L$, and its witnessing enumeration belongs to the ambient universe too. Thus every such ordinal is countable externally, proving ${\omega }_1^L\le {\omega }_1$. Consequently the ambient cardinality of the constructible reals is at most ${\aleph }_1$.

If every subset of $\omega$ were constructible, the displayed bijection would enumerate the full ambient power set using at most ${\aleph }_1$ points. This contradicts the assumption $2^{{\aleph }_0}={\aleph }_2$. ThereforeThe argument permits ${\omega }_1^L$ to be strictly smaller than the ambient ${\omega }_1$, exactly the cardinal-absoluteness issue highlighted in the source.