For $Ly=-\lambda y$, self-adjointness gives $-\lambda \Vert y{\Vert }^2=⟨y,Ly⟩=\overline{⟨y,Ly⟩}$. Since an eigenfunction is nonzero, $\boxed{\lambda \in \mathbb{R}}$. One may also see reality directly from $\lambda \int |y{|}^2=\int |y^{\prime }{|}^2-\int |y{|}^2$, whose boundary term vanishes under the mixed conditions.

Assume a nonempty set $a$ has no membership-minimal element, so for every $x\in a$ there is $y\in x\cap a$. Consider the formula $\phi (x)\equiv x\notin a$.

It is progressive: if every element of $x$ is outside $a$ but $x\in a$, the assumed absence of a minimal element supplies $y\in x\cap a$, contradicting that hypothesis. The membership induction theorem therefore implies $x\notin a$ for every set $x$. This contradicts $a\ne \varnothing$.

Thus every nonempty set $a$ contains an element $x$ with $x\cap a=\varnothing$, exactly the Axiom of foundation:Only the induction schema and elementary set reasoning were needed; no prior use of Axiom of foundation occurs in the proof.

The cumulative hierarchy is defined by transfinite recursion:This recursion on ordinals is available without assuming the Axiom of foundation for all sets. The stages are increasing and transitive, by transfinite induction using their definitions.

Work internally in the specified model of ZF without the Axiom of foundation, and assume the membership induction theorem. Apply it toSuppose every member $y$ of $x$ lies in some stage. Assign to each such $y$ the least ordinal ${\alpha }_y$ for which $y\in V_{{\alpha }_y}$. The Axiom schema of replacement gives a set of these ordinals, and let $\beta$ be their supremum, taking $\beta =0$ for the empty collection. Monotonicity gives $x\subseteq V_{\beta }$, and therefore $x\in V_{\beta +1}$.

This proves progressivity of $\phi$, so membership induction gives it for every set of the model. Every stage already consists of sets in that model. ConsequentlyThe equality is internal to the given model; its ordinals and power sets are computed there.

For $\kappa ={\aleph }_{\omega }$, monotonicity of cardinal exponentiation givesThe middle equality uses the currying law for cardinal exponentiation, and the final one uses ${\aleph }_0\cdot \kappa =\kappa$. Hence ${\aleph }_0^{\kappa }=2^{\kappa }$, even before assuming GCH. Applying the Generalized continuum hypothesis yields

Put $\kappa ={\aleph }_{\omega }$. Apply König theorem for cardinal numbers to the sequences ${\aleph }_n<{\aleph }_{n+1}$:The sum is $\kappa$ because the countable union has cardinality at most $\kappa$ and its summands are cofinal in $\kappa$. Thus the power is strictly larger than $\kappa$, a fact that cannot be replaced by the regular-cardinal calculation of the preceding part.

For an upper bound, use $\kappa \le 2^{\kappa }$ and the currying law for cardinal exponentiation:where the last step is GCH at $\kappa$. There is no cardinal strictly between $\kappa$ and its successor, so

Constant functions give the lower bound ${\aleph }_2\le {\aleph }_2^{{\aleph }_0}$. By the Generalized continuum hypothesis, ${\aleph }_2=2^{{\aleph }_1}$. The currying law for cardinal exponentiation and infinite cardinal arithmetic therefore giveHence $\boxed{{\aleph }_2^{{\aleph }_0}={\aleph }_2}$.

A bounded expression for the successor relation isThe first two clauses put $y$ and all its elements into $x$, while the last forbids any additional member of $x$. Extensionality identifies this exactly with $x=y\cup \{y\}$. Since the formula is ${\Delta }_0$, it is absolute in both directions between the transitive classes and therefore upward absolute. This bounded description also justifies the successor computations used earlier in the rank-stage Replacement counterexample.

Equality with the union is expressed byThe first conjunct says that every member of $x$ lies in some member of $y$; the second gives the reverse inclusion. Every quantifier is bounded by a parameter or by an element of a previously bounded parameter. Therefore bounded absoluteness applies, proving full absoluteness and hence upward absoluteness of $x=\bigcup y$. This only compares the relation for existing parameters; it does not require either class to contain the union of every one of its sets.

Use the bounded formulaIt expresses $x\subseteq y$ with no unbounded quantifier. For parameters $x,y\in U_1$, transitivity puts every member of $x$ in $U_1$ already, and membership in $y$ has the same meaning in both classes. Thus the formula is ${\Delta }_0$ and is absolute in both directions, in particular upward absolute.

The product from the preceding part injects into ${\kappa }^{{\aleph }_0}$, where $\kappa ={\aleph }_{\omega }$, because each coordinate value is an ordinal below $\kappa$. Combining its strict König lower bound with the exponentiation upper bound givesHenceOnly the instance $2^{{\aleph }_{\omega }}={\aleph }_{\omega +1}$ of the generalized continuum hypothesis was needed in these three computations, though the question assumes the full hypothesis. No assumption that countable exponentiation leaves this singular cardinal unchanged is valid: König's theorem rules that out.

Write again $\kappa ={\aleph }_{\omega }$. The countable cardinal sum ${\sum }_{n<\omega }{\aleph }_n$ equals $\kappa$: it is at least their supremum, and at most ${\aleph }_0\cdot \kappa =\kappa$. König's theorem givesThe latter product injects into $P={\prod }_{n<\omega }{\aleph }_n$ by fixing coordinate zero and putting its $n$th coordinate in coordinate $n+1$. Conversely $P$ injects into the set of all functions $\omega \to \kappa$. ThereforeNo cardinal lies strictly between $\kappa$ and its successor, soThe strict lower bound is essential: replacing the product by the supremum of its factors would incorrectly give ${\aleph }_{\omega }$.

For GCH arithmetic at aleph omega, let $\kappa ={\aleph }_{\omega }$. Use monotonicity of cardinal exponentiation, the law $(a^b{)}^c=a^{bc}$, and $\kappa \cdot \kappa =\kappa$ for infinite cardinals in ZFC. Since $2\le \kappa \le 2^{\kappa }$,Thus ${\kappa }^{\kappa }=2^{\kappa }$, and the generalized continuum hypothesis yieldsThe lower bound embeds binary-valued functions into $\kappa$-valued ones; the upper bound codes each value by a subset and combines the resulting pairs of indices. In particular the equality before applying the hypothesis is a ZFC equality.

Induct on $\alpha$. For every $\beta <\alpha$, the induction hypothesis says $\beta \in V_{\beta +1}$, and monotonicity gives $V_{\beta +1}\subseteq V_{\alpha }$. Thus every element of the ordinal $\alpha$ lies in $V_{\alpha }$, so $\alpha \subseteq V_{\alpha }$. Taking its power set givesFor $\alpha =0$ the inclusion is vacuous and $\varnothing \in \mathcal{P}(\varnothing )$ supplies the base case. This argument uses ordinal membership as strict order and the previously proved hierarchy monotonicity.

For $x\in V_{\alpha }$, transitivity of $V_{\alpha }$ gives $x\subseteq V_{\alpha }$. Therefore $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$, provingTogether with the union definition at limits, transfinite induction now gives $V_{\alpha }\subseteq V_{\beta }$ whenever $\alpha \le \beta$. No rank theorem is needed for this monotonicity argument.

A set $T$ is transitive if every element of an element of $T$ is itself in $T$, equivalently each element of $T$ is a subset of $T$. Proceed by transfinite induction. The empty set is transitive. If $T$ is transitive and $u\in v\in \mathcal{P}(T)$, then $v\subseteq T$ gives $u\in T$, and transitivity of $T$ gives $u\subseteq T$. Thus $u\in \mathcal{P}(T)$, proving the successor stage transitive. A union of transitive sets is transitive: if $u\in v\in {\bigcup }_i{T}_i$, choose $i$ with $v\in T_i$ and infer $u\in T_i$. This proves the limit step and hence every $V_{\alpha }$ is transitive.

A ${\Sigma }_1$ logical formula has the formup to the standard logical transformations defining the hierarchy. The initial existential quantifiers may be unbounded; all quantifiers inside the matrix are bounded. A finite tuple of witnesses can be encoded by one set if desired, and logical formulas with no initial unbounded quantifier are included as the level-zero special case.

The dual ${\Pi }_1$ class in set theory has an initial universal block with a bounded matrix. Negation interchanges these two classes, but does not in general preserve the existential class in set theory. Part (iii) supplies an explicit reason for the failure in this setting.

A bounded quantifier has the form $\exists u\in v$ or $\forall u\in v$, abbreviating $\exists u(u\in v\wedge \cdots )$ or $\forall u(u\in v\to \cdots )$. A ${\Sigma }_0$ logical formula, also called ${\Delta }_0$, is built from atomic membership and equality logical formulas using Boolean connectives and only bounded quantifiers. It has no unbounded quantifier.

For example $z\subseteq y$ abbreviates the bounded logical formula $\forall u\in z(u\in y)$. Quantifiers are bounded by an existing variable or parameter, not by an unrestricted class term such as the power set of a parameter unless that set has separately been supplied as a parameter.

A topological space is Lindelöf if every open cover has a countable subcover. We also prove the requested equivalence for metrizable spaces here, after the two source definition leaves.

Choose a compatible metric $d$. First suppose that $D$ is a countable dense subset. The balls $B(d_0,q)$ with centers $d_0\in D$ and positive rational radii $q$ form a countable topological basis. Indeed, if $x\in U$ and $B(x,ϵ)\subset U$, choose $d_0\in D$ with $d(x,d_0)<ϵ/3$ and a rational $q$ between $d(x,d_0)$ and $ϵ-d(x,d_0)$. The triangle inequality gives $x\in B(d_0,q)\subset U$.

For any open cover, retain the topological basis elements which are contained in some member of the cover, and choose one covering member for each such topological basis element. There are only countably many choices, and the resulting subfamily covers the topological space, since each point has a topological basis neighborhood inside a covering member. Thus a separable metrizable space is Lindelöf.

Conversely, suppose the metric space is Lindelöf. For each positive integer $n$, the balls $B(x,1/n)$ cover it, so select a countable subcover with centers in a countable set $D_n$. The union $D={\bigcup }_{n\ge 1}{D}_n$ is countable and dense: if $B(x,ϵ)$ is a nonempty open ball, choose $n$ with $1/n<ϵ$. One of the selected radius-$1/n$ balls contains $x$, so its center lies in $D\cap B(x,ϵ)$. ThereforeThe metric hypothesis is used in producing uniform small-ball covers and a countable ball topological basis; the equivalence need not hold for general topological spaces.

A topological space is separable if it has a countable dense subset. A subset $D$ is dense when its closure is the whole topological space, equivalently when every nonempty open set intersects $D$. Countable here permits a finite set, so finite topological spaces and the empty topological space fit the same definition.

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.