There are two ways for a set to fail to be the power set of $y$: it may contain an object which is not a subset of $y$, or omit a subset of $y$. A single existential logical formula expressing these alternatives isIts matrix is bounded, so this is ${\Sigma }_1$. Its negation is precisely the first-order power-set specification

We prove the absoluteness fact needed to exclude a logically equivalent ${\Sigma }_1$ logical formula. If $M\subseteq N$ are transitive sets and all parameters lie in $M$, bounded logical formulas have the same truth value in $⟨M,\in ⟩$ and $⟨N,\in ⟩$. Atomic equality and membership agree because both structures use actual membership. Boolean operations preserve agreement. In a bounded quantifier over $v\in M$, transitivity ensures that every element of $v$ already lies in $M$, so the two structures quantify over exactly the same elements of $v$. Induction on logical formulas proves the claim. It follows that ${\Sigma }_1$ logical formulas are upward absolute: witnesses in $M$ remain available in $N$, and their bounded matrix retains its truth value.

Now use a finite counterexample, so no existence of a transitive model of ZF is assumed. Let $M=4=\{0,1,2,3\}$, the Finite von Neumann ordinal, and let $N=V_4$. These are transitive sets with $M\subseteq N$. Set $y=2$ and $x=3$, both elements of $M$. Inside $M$, the subsets of $2$ which are available as elements of the domain are exactly $0,1,2$; the remaining domain element $3$ is not a subset of $2$. Thus $M$ satisfies the displayed power-set specification with $x=3$.

Inside $N$, however, the set $\{1\}$ is also present and is a subset of $2$, while $\{1\}\notin 3$. Thus the same specification is false in $N$. If $\neg \varphi$ were logically equivalent to a ${\Sigma }_1$ logical formula, its truth in $M$ would transfer upward to $N$, a contradiction. ThereforeThis proves the requested unrestricted logical non-equivalence. The finite structures are not being claimed to satisfy all of ZF; logical equivalence must hold in them as well as in any models of a stronger theory.

Use the strict-order convention appropriate to the membership relation. A bounded logical formula isThe first clause is irreflexivity, the second transitivity, and the third comparability of distinct elements. Irreflexivity and transitivity exclude $u\in v\in u$, so distinct elements cannot be ordered in both directions. These are exactly the axioms of a strict total order on the set $x$, with $\in$ as its order relation.

Every displayed quantifier is bounded by $x$, so $TO(x)$ is ${\Sigma }_0$. The logical formula does not assert that $x$ is transitive as a set, and no such additional assertion is needed to say that the restriction of membership to $x$ is a total order.

Fix a logical formula $\theta (x,y,\vec{p})$ with parameters in $V$, and a set $a\in V$. Suppose $⟨V,\in ⟩$ satisfies that each $x\in a$ has a unique $y$ with $\theta (x,y,\vec{p})$. Relativize every unbounded quantifier in $\theta$ to the definable class in set theory $V$, obtaining an ambient first-order formula ${\theta }^V$.

Transitivity of $V$ ensures that every member of $a$ belongs to $V$. The ambient relationtherefore defines a unique ambient set $y$ for every $x\in a$. Apply ambient replacement to obtain its range set $b$. By construction every element of $b$ belongs to $V$, so part (ii) gives $b\in V$.

Membership in this set $b$ is exactly the range relation required by replacement inside the structure $⟨V,\in ⟩$. Since the logical formula and parameters were arbitrary, every instance of the replacement scheme holds in the hierarchy class in set theory. The proof uses formula relativization to a class and the closure property from part (ii), not the unsupported assertion that $V$ is the whole ambient universe.

For every $x\in a$, the assumption $a\subseteq V$ supplies an ordinal stage containing $x$. Let ${\alpha }_x$ be the least such ordinal. Leastness makes this a uniquely definable assignment, so ambient replacement gives the setPut $\beta =\sup S$, taking $\beta =0$ if $a$ is empty. Monotonicity of the hierarchy gives $x\in V_{{\alpha }_x}\subseteq V_{\beta }$ for every $x\in a$. Hence $a\subseteq V_{\beta }$, and thereforeThis is a rank-bounding argument: a set-sized collection of elements from the hierarchy has its stages bounded by one ordinal. No simultaneous arbitrary choices are required, since each least stage is uniquely specified.

In first-order set theory a class in set theory is a definable collection, written schematically $\{x:\theta (x,\vec{p})\}$ using a logical formula and set parameters. A class term is an abbreviation for such a defining condition; it does not add a new kind of object to the quantifiers of ZF. A class in set theory is proper if it is not the collection of elements of any set.

Define the cumulative hierarchy by transfinite recursion:and let $V=\{x:\exists \alpha \text{ ordinal }(x\in V_{\alpha })\}$. Without foundation this denotes the well-founded part produced by the hierarchy; it must not simply be assumed to contain every ambient set. Transfinite recursion on ordinals and the relevant set constructions remain available in ZF without foundation.

Prove transitivity by transfinite induction. The empty stage is transitive. Suppose $T=V_{\alpha }$ is transitive. Then each member of $T$ is a subset of $T$, so $T\subseteq \mathcal{P}(T)$. Also $\mathcal{P}(T)$ is transitive: if $u\in v\in \mathcal{P}(T)$, then $u\in T$ and, by transitivity of $T$, $u\subseteq T$, so $u\in \mathcal{P}(T)$. At a limit stage, a union of transitive sets is transitive. This proves transitivity at every stage.

Finally transitivity of $V_{\alpha }$ itself implies that every member is a subset of $V_{\alpha }$, givingThese successive inclusions and the limit definition also imply $V_{\alpha }\subseteq V_{\beta }$ whenever $\alpha \le \beta$. The argument has not used foundation. This leaf includes the unnumbered definitions preceding the source's three claims.

Use the Stone-Čech extension property: a discrete topological space $X$ embeds densely into a compact Hausdorff space $\beta X$, and every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Apply it to the characteristic function ${\chi }_A:X\to \{0,1\}$ of an arbitrary subset $A$; discreteness makes it continuous. Its extension ${\tilde{\chi }}_A$ has clopen fibers, andIndeed that fiber is closed and contains $A$, while its openness and density of $X$ show that every point in the fiber is approached by points of $A$.

If $A$ and $B$ are disjoint, the extension of ${\chi }_A$ is one on $\overline{A}$ and zero on $\overline{B}$, so these closures are disjoint. The inclusion $X\to \beta X$ therefore satisfies the criterion from part (b), and its ultrafilter-limit map is both surjective and injective. ThusThis is a bijection of underlying sets; identifying the compatible topology on the ultrafilter topological space would give the familiar stronger homeomorphism statement, but is not needed for the requested conclusion.

For every ultrafilter $\mathcal{H}$ on $X$, its image is an ultrafilter on the compact space $Y$, so it has a limit. That limit is unique because $Y$ is Hausdorff: distinct limits would have disjoint neighborhoods both belonging to the same proper filter. Therefore $\Lambda$ is well-defined. Since $f[X]$ is dense, every $b\in Y$ lies in $\overline{f[X]}$. Part (a), with $A=X$, supplies an ultrafilter mapping to $b$, proving surjectivity.

Suppose disjoint subsets have disjoint image closures. If $\mathcal{H}\ne \mathcal{K}$, choose $A\in \mathcal{H}$ not in $\mathcal{K}$; the ultrafilter criterion gives $X\setminus A\in \mathcal{K}$. Part (a) givesThe two closures are disjoint by assumption, so the limits differ. Thus $\Lambda$ is injective.

Conversely, suppose $\Lambda$ is injective and disjoint $A,B\subseteq X$ had a common point $b$ in their image closures. Part (a) would supply ultrafilters $\mathcal{H},\mathcal{K}$ containing $A,B$ respectively, both with image limit $b$. They must be distinct, since a proper filter cannot contain the disjoint sets $A$ and $B$ simultaneously. This contradicts injectivity. Hence

A proper filter $\mathcal{F}$ on $X$ is a nonempty family of subsets, closed under finite intersection and passage to supersets, with $\varnothing \notin \mathcal{F}$. An ultrafilter is a maximal proper filter, equivalently one which contains, for each $A\subseteq X$, exactly one of $A$ and $X\setminus A$. The pushforward filter isInverse images preserve intersections and complements. Thus it is a proper filter, and the either-a-set-or-its-complement criterion shows it is an ultrafilter whenever $\mathcal{F}$ is one.

Convergence of a filter $\mathcal{F}\to x$ means that every neighborhood of $x$ belongs to $\mathcal{F}$. A mapping is continuous exactly when $\mathcal{F}\to x$ implies $f_{\ast }\mathcal{F}\to f(x)$ for all proper filters and points. The forward implication follows from inverse images of neighborhoods. For the converse apply the criterion to the neighborhood proper filter at $x$: each inverse image of a neighborhood of $f(x)$ must be a neighborhood of $x$, which is continuity at $x$.

A topological space is compact exactly when every ultrafilter converges to at least one point; equivalently every proper filter has a cluster point. For completeness, in a compact space the closed sets $\overline{A}$, $A\in \mathcal{U}$, have the finite intersection property, hence a common point $x$. Every neighborhood $N$ of $x$ belongs to the ultrafilter. Indeed, if $N$ did not, choose an open $V$ with $x\in V\subseteq N$. Then $V$ would not belong either, so $X\setminus V$ would belong. That set is closed and excludes $x$, contradicting the choice of $x$ in every closure. Conversely, if an open cover has no finite subcover, its closed complements generate a proper filter. An ultrafilter extension cannot converge anywhere, because each point has a covering neighborhood whose complement belongs to that ultrafilter. This proves the compactness criterion using the permitted extension principle.

Now suppose $b\in \overline{f[A]}$. The familyis a filter base of nonempty sets: closure guarantees each intersection is nonempty, and finite intersections correspond to intersections of neighborhoods. Extend its generated proper filter to an ultrafilter $\mathcal{H}$. It contains $A$ and every $f^{-1}[V]$, so $f_{\ast }\mathcal{H}\to b$.

Conversely, if such an ultrafilter exists, each neighborhood $V$ of $b$ has $f^{-1}[V]\in \mathcal{H}$. Its intersection with $A\in \mathcal{H}$ is nonempty. Thus $V$ meets $f[A]$, provingThis part includes the unnumbered proper filter definitions and criteria preceding the printed subparts.

First prove the suggested density fact. If $Z$ is dense in $Y$, $V$ is open and $y\in V$, then every open neighborhood $W$ of $y$ has $W\cap V$ nonempty open. It meets $Z$, so $W$ meets $V\cap Z$. HenceIdentify $X$ with its dense embedded image in the hausdorff compactification. For $x\in X$, local compactness supplies an open neighborhood $U$ in $X$ and a compact set $K\subseteq X$ with $x\in U\subseteq K$. Choose an open set $V$ in $Y$ with $V\cap X=U$. The inclusion into $Y$ is continuous, so $K$ is compact in $Y$ and therefore closed, since $Y$ is Hausdorff. The density fact now yieldsThus every point of the embedded $X$ has a $Y$-open neighborhood contained in $X$. Consequently

A Hausdorff compactification of $X$ is a pair $(h,Y)$ in which $Y$ is compact Hausdorff and $h:X\to Y$ is a topological embedding with dense image.

For a locally compact Hausdorff noncompact topological space, form $X^{\ast }=X\cup \{\infty \}$ with one new point. Keep all open subsets of $X$, and take the neighborhoods of infinity to contain setsCompact subsets of a Hausdorff space are closed. Finite unions of compact sets are compact, so finite intersections of these basic neighborhoods are again of the same form; their intersections with ordinary open sets are open in $X$. This specifies a topology restricting to the original one on $X$.

It is compact: from an open cover choose a member containing infinity. Its complement is contained in some compact $K$, and finitely many more members cover that compact remainder. It is Hausdorff: two points of $X$ are separated as before; to separate $x\in X$ from infinity, choose an open neighborhood $U$ of $x$ lying inside a compact neighborhood $K$. The open set $U$ and the infinity neighborhood $X^{\ast }\setminus K$ are disjoint.

The inclusion of $X$ is an embedding. Every basic neighborhood of infinity meets $X$, because otherwise its compact complement $K$ would equal all of $X$, contrary to noncompactness. Thus $X$ is dense in $X^{\ast }$. This proves that $X^{\ast }$ is the one-point compactification.

Let $\mathcal{B}$ be a topological basis consisting of clopen sets. For $U\in \mathcal{B}$ its characteristic function ${\chi }_U:X\to \{0,1\}$ is continuous, since the preimages of both singleton values are open. DefineThe coordinate criterion for product topology makes this map continuous. If $x\ne y$, the $T_0$ property gives an open set containing one and not the other. A topological basis element contained in that open set still distinguishes them, so their corresponding characteristic coordinates differ. Therefore $e$ is injective.

Finally, for each topological basis element $U$,The right-hand side is the preimage of an open cylinder in the product of discrete two-point topological spaces. These cylinders recover a topological basis for the topology of $X$, proving continuity of the inverse onto the image. Hence

Urysohn's lemma states that, in a normal space, disjoint closed subsets $A,B$ can be separated by a continuous function $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

First use the shrinking consequence of the separation property of a normal space: if a closed set $F$ lies in an open set $G$, there is an open $W$ with $F\subseteq W\subseteq \overline{W}\subseteq G$. To prove it, separate the disjoint closed sets $F$ and $X\setminus G$ by disjoint open neighborhoods $W,V$. Then $\overline{W}\subseteq X\setminus V\subseteq G$.

Let $D$ be the dyadic rationals in $[0,1]$. Set $U_1=X\setminus B$, and shrink around $A$ to choose $U_0$ with $A\subseteq U_0\subseteq \overline{U_0}\subseteq U_1$. At each subsequent dyadic stage, insert between each adjacent pair $s<t$ an open set $U_{(s+t)/2}$ such thatThe shrinking lemma applies because the first set is closed and contained in the last open set. Induction gives $\overline{U_s}\subseteq U_t$ for all dyadic $s<t$. Part (a) now constructs a continuous $u$. Each point of $A$ lies in $U_0$, so has value zero. Each point of $B$ lies in none of the $U_t$, so has value one. This proves the lemma.

For the embedding, let $I=C(X,[0,1])$, a set of continuous functions, and define the evaluation mapEvery coordinate is continuous, so $e$ is continuous in the product topology. The Hausdorff assumption makes singletons closed. For distinct points, Urysohn's lemma applied to their singletons gives a coordinate with different values, so $e$ is injective.

To check the inverse topology, take $x\in O$ with $O$ open. Apply the lemma to $\{x\}$ and $X\setminus O$, choosing a coordinate $u$ with $u(x)=1$ and $u=0$ outside $O$. ThenThis is the preimage of an open coordinate cylinder under $e$. Thus every original neighborhood is detected by coordinate neighborhoods, and $e^{-1}:e[X]\to X$ is continuous. The evaluation map is a topological embedding.

For each point which belongs to at least one $U_t$, its membership-index set is a nonempty subset of $[0,1]$, so has an infimum in $[0,1]$. Assigning the value one to the other points therefore defines $f:X\to [0,1]$ everywhere.

For $0<r\le 1$,An index $t<r$ gives $f(x)\le t<r$. Conversely if the infimum is below $r$, some membership index must be below $r$. Thus this strict sublevel set is open.

For $0\le r<1$ we claimIf $x\notin \overline{U_t}$ and $t>r$, then $x$ cannot belong to any $U_s$ with $s\le t$, since $U_s\subseteq U_t$ for $s<t$. Hence $f(x)\ge t>r$. Conversely, if $f(x)>r$, density of $D$ supplies $r<t<u<f(x)$ with $t,u\in D$. Since $x\notin U_u$ and $\overline{U_t}\subseteq U_u$, one has $x\notin \overline{U_t}$. This proves the identity, and its right-hand side is open.

The remaining sublevel and superlevel sets, for thresholds outside these ranges, are either empty or all of $X$. Preimages of open rays are consequently open, and these rays generate the usual real topology. Therefore $f$ is continuous. The closure inclusion is what supplies openness of the strict superlevel sets; mere nesting would not suffice.

The lower limit topology on the real line is the lower limit topology. Every nonempty basic interval $[a,b)$ contains a rational number, so the rational numbers are a countable dense subset. By the preceding product argument, ${\mathbb{Q}}^2$ is dense in the square, proving its topological separability.

Consider the uncountable anti-diagonalIt is a discrete subset: the basic rectangle $[t,t+ϵ)\times [-t,-t+ϵ)$ intersects $D$ only at $(t,-t)$, because both coordinates have increased weakly and their sum can remain zero only if neither increases.

It is also closed. At a point $(u,v)$ off the anti-diagonal, $u+v\ne 0$. If this sum is positive, every sufficiently small lower-limit rectangle based at $(u,v)$ still has positive coordinate sum. If it is negative, choose $ϵ>0$ with $2ϵ<-(u+v)$; then $[u,u+ϵ)\times [v,v+ϵ)$ has negative coordinate sum throughout. These neighborhoods avoid $D$, so its complement is open.

If the square were Lindelöf, its closed discrete subset $D$ would be countable by part (c), contradicting its bijection with $\mathbb{R}$. Hence

Since $D$ is a discrete subset in the subspace topology, for each $d\in D$ choose an open set $U_d\subseteq Z$ with $U_d\cap D=\{d\}$. Closedness of $D$ makes $Z\setminus D$ open. The familyis therefore an open cover of $Z$. A countable subcover exists by the Lindelöf property. Each point $d$ can be covered only by its own $U_d$ among this family: the complement misses all of $D$, and every other $U_e$ meets $D$ only at $e$. Hence the countable subcover must include $U_d$ for every $d\in D$. This forcesBoth closedness and discreteness matter: they provide the open complement and the single-point traces, respectively.

Choose countable dense sets $D_X\subseteq X$ and $D_Y\subseteq Y$. Their Cartesian product is countable. Every nonempty basic open set of the product topology has the form $U\times V$ with $U,V$ nonempty open in their respective topological spaces. Density gives $d_X\in D_X\cap U$ and $d_Y\in D_Y\cap V$, hence $(d_X,d_Y)\in (D_X\times D_Y)\cap (U\times V)$.

Every nonempty open set in the product contains such a basic open set. Thus $D_X\times D_Y$ is dense in $X\times Y$, proving topological separability of the product. If one factor is empty the product is empty and the conclusion holds directly.