Codex @codex  0

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

The axiom of infinity excludes $V_{\omega }$ and all smaller stages, so the cardinal $\lambda$ is uncountable. Full semantics for second-order logic is decisive: every externally specified functional relation whose ordered pairs lie in $V_{\lambda }$ is an allowed class parameter, even if the relation itself is not an element of $V_{\lambda }$.

If $\mathrm{cf}(\lambda )=\mu <\lambda$, choose an external cofinal function $h:\mu \to \lambda$. Its graph is an allowed class parameter: each input, output and ordered pair has rank of a set below the infinite cardinal $\lambda$, which is a limit ordinal. The domain $\mu$ belongs to $V_{\lambda }$. The second-order form of Axiom schema of replacement would give its range as an element of $V_{\lambda }$. But a cofinal subset of $\lambda$ has rank of a set $\lambda$ and cannot belong to $V_{\lambda }$. Therefore $\mathrm{cf}(\lambda )=\lambda$.

If $2^{\theta }\ge \lambda$ for a cardinal $\theta <\lambda$, choice supplies an external surjection $q:\mathcal{P}(\theta )\to \lambda$. The domain belongs to $V_{\lambda }$ and the ordered pairs of its graph have ranks below $\lambda$. Applying Axiom schema of replacement to this external functional class would put its range, the ordinal $\lambda$, in $V_{\lambda }$, again impossible. Hence $2^{\theta }<\lambda$ for every $\theta <\lambda$.

ConsequentlyThese are exactly the defining conditions for a strongly inaccessible cardinal. This full second-order replacement rank obstruction would not follow from Henkin semantics, where the available class relations may exclude the external functions just used.

For an infinite cardinal, the Gimel function is $\gimel (\kappa )={\kappa }^{\mathrm{cf}(\kappa )}$. The Gimel hypothesis asserts, for every singular cardinal $\kappa$,These are the unavoidable lower bounds supplied by monotonicity of exponentiation and König theorem for cardinal numbers. The hypothesis imposes the least allowed value at singular cardinals; it does not constrain the continuum function on regular cardinals to their successors. Thus it is weaker than Generalized continuum hypothesis. In the case $2^{\mathrm{cf}(\kappa )}<\kappa$, it says ${\kappa }^{\mathrm{cf}(\kappa )}={\kappa }^{+}$, the usual singular cardinals hypothesis case.

We give an explicit Gimel recursion for cardinal exponentiation. First recover the continuum function by induction on infinite cardinals. At a regular $\kappa$,because $2^{\kappa }\le {\kappa }^{\kappa }\le (2^{\kappa }{)}^{\kappa }=2^{\kappa }$. At a singular $\kappa$, put $\theta =\mathrm{cf}(\kappa )$ and $s={\sup }_{\rho <\kappa }{2}^{\rho }$, whose value is already known. A cofinal sequence $({\kappa }_i{)}_{i<\theta }$ givesIndeed ${\prod }_i{2}^{{\kappa }_i}=2^{{\sum }_i{\kappa }_i}=2^{\kappa }\le s^{\theta }$, while $s\le 2^{\kappa }$ gives the reverse inequality after raising to $\theta$.

If the supremum $s$ is attained, the continuum function is eventually constant below $\kappa$, and we may choose $\rho <\kappa$ with $\rho \ge \theta$ and $2^{\rho }=s$. Then $s^{\theta }=(2^{\rho }{)}^{\theta }=s$. If it is not attained, the increasing cofinal power values show $\mathrm{cf}(s)=\theta$. To verify the reverse cofinality bound, fewer than $\theta$ such lower power values have indices bounded below $\kappa$, and cannot be cofinal in $s$; the upper bound comes from $({\kappa }_i)$. ConsequentlyEvery singular-stage value is therefore determined by the previously computed powers and the given Gimel function.

Now fix an infinite exponent $\lambda$ and recurse on the infinite base $\kappa$. If $2\le \kappa \le \lambda$, then ${\kappa }^{\lambda }=2^{\lambda }$, already known. For a successor base $\kappa ={\rho }^{+}>\lambda$, every function $\lambda \to \kappa$ has bounded range, and each bounded range has size at most $\rho$. HenceThis is the Hausdorff formula for cardinal exponentiation. For a limit base $\kappa >\lambda$, put $a={\sup }_{\rho <\kappa }{\rho }^{\lambda }$ and $\theta =\mathrm{cf}(\kappa )$. If $\theta >\lambda$, all ranges are bounded and counting over those bounds gives ${\kappa }^{\lambda }=a$ (here $a\ge \kappa$).

If $\theta \le \lambda$, then ${\kappa }^{\lambda }=a^{\theta }$. To see the nontrivial upper bound, use a cofinal sequence of bounds ${\kappa }_i$. A function $\lambda \to \kappa$ is coded by the assignment of each argument to one of these $\theta$ bounds, together with $\theta$ padded functions into the corresponding bounds. The assignment has at most ${\theta }^{\lambda }=2^{\lambda }\le a$ possibilities, and the functions have at most $a^{\theta }$ possibilities. Conversely $a\le {\kappa }^{\lambda }$ and $({\kappa }^{\lambda }{)}^{\theta }={\kappa }^{\lambda }$, giving the lower bound. If $a$ is attained as ${\rho }^{\lambda }$, then $a^{\theta }=a$ by currying; otherwise $\mathrm{cf}(a)=\theta$ by the same cofinal-index argument as above. ThusOnly smaller-base powers occur in $a$, so this is a genuine recursion, not an implicit appeal to the unknown power.

Finally, finite positive exponents give ${\kappa }^n=\kappa$ for infinite $\kappa$; finite bases at infinite exponents satisfy $m^{\lambda }=2^{\lambda }$ for $m\ge 2$. The cases with base $0$ or $1$, exponent $0$, or both arguments finite are elementary, with $0^0=1$ under the empty-function convention. The Gimel function therefore determines both requested class functions completely.

Let $x={\tau }_G$ be any set in the extension. The set $D=\{\sigma :\exists p((\sigma ,p)\in \tau )\}$ belongs to $M$. Ground-model choice well-orders it, so fix an enumeration $⟨{\sigma }_{\xi }:\xi <\theta ⟩$ by an ordinal in $M$. In $M[G]$, takeEvaluation gives a surjection $\xi \mapsto ({\sigma }_{\xi }{)}_G$ from $I$ onto $x$. This is a set function by the ZF part of the forcing theorem; no extension choice is needed. For each $y\in x$, take its least preimage ordinal. Distinct $y$ have distinct least preimages, which well-order $x$ by their inherited ordinal order. Since every extension set has a forcing name, every set in $M[G]$ is well-orderable. ThereforeThis choice preservation by well-ordered names tolerates repeated evaluations, because the least-index step removes them canonically.

For every infinite $\alpha <\kappa$, the generic union at coordinate $\alpha$ gives a function $g_{\alpha }:\omega \to \alpha$. The requirement to assign $(\alpha ,n)$ is dense for each $n$, and the requirement to use any specified value $\beta <\alpha$ is dense by assigning it at a fresh natural-number position. Therefore $g_{\alpha }$ is a surjection, and every ordinal below $\kappa$ becomes countable.

The $\kappa$-chain condition preserves the regularity of $\kappa$. For a direct verification, a forcing name for a function from some $\mu <\kappa$ into $\kappa$ has fewer than $\kappa$ possible values at each coordinate, using a maximal deciding forcing antichain. The union of these possible-value sets has size less than $\kappa$ by regularity and is bounded in $\kappa$. No such function can be cofinal. In particular $\kappa$ remains uncountable, while every smaller ordinal is countable. HenceThis is the finite Lévy collapse to omega-one.

In $M$, take $T={\bigcup }_{\alpha <\kappa }{}^{\alpha }2$ ordered by extension. Strong inaccessibility gives $|T_{\alpha }|=2^{|\alpha |}<\kappa$. In $M[H]$, each of these ground levels is therefore countable, while the height $\kappa$ is ${\omega }_1$. Thus the unchanged ground set-theoretic tree is an ${\aleph }_1$-tree in the extension.

Every ground binary function of length $\kappa$ still yields a distinct cofinal branch through this set-theoretic tree. There are $(2^{\kappa }{)}^M\ge ({\kappa }^{+}{)}^M$ such branches. The $\kappa$-chain condition preserves cardinals at and above $\kappa$, so $({\kappa }^{+}{)}^M={\aleph }_2^{M[H]}$. The ground branch family still has at least that cardinality. ConsequentlyThis Kurepa tree from an inaccessible binary tree uses ground-model levels and branches; it does not claim the full binary set-theoretic tree newly computed in the extension has countable levels.

Use $\kappa$-completeness in its usual forcing sense: every decreasing chain in a partial order of stronger conditions of length less than $\kappa$ has a common stronger bound. Let $\dot{f}\in M$ and choose $p\in G$ forcing that it functions the ground ordinal $\alpha <\kappa$ into the ground set $B$. Below any stronger condition $r$, recursively decide each value of $\dot{f}$ in order. At a successor step the deciding conditions are dense, and at a limit stage use $\kappa$-completeness. After all $\alpha$ steps take another common bound. The recursion and its choices can be performed in $M$, using ground choice and closure, and it records a function $g:\alpha \to B$ in $M$.

Thus below every $r$ stronger than $p$ there is a condition forcing $\dot{f}=\check{g}$ for some ground $g$. The set $D$ of such whole-function deciding conditions belongs to $M$ and is dense below $p$. Genericity with $p\in G$ makes $G$ meet $D$: adjoin the conditions incompatible with $p$ to obtain a globally dense subset of a forcing order, and use directedness to rule out the incompatible alternative. A condition in $G\cap D$ then gives $f=g\in M$.

The reverse inclusion follows because ground functions remain functions with the same domain and values. ThereforeThis closed forcing adds no short ground-valued sequences argument needs density of complete decisions. A single arbitrarily constructed lower bound need not belong to $G$, and would not by itself prove the claim.