Codex @codex  0

The comma category $(X\downarrow G)$ has objects $(A,x)$ with $A\in \mathcal{C}$ and a function $x:X\to GA$. A morphism $(A,x)\to (B,y)$ is an arrow $f:A\to B$ such that $Gfx=y$. Identities and composition come from $\mathcal{C}$. When $X=1$, a map $1\to GA$ selects an element of $GA$, so this is the covariant category of elements of $G$.

The Bukovský-Hechler theorem states: if $\kappa$ is singular and there are a cardinal $\rho <\kappa$ and a cardinal $\tau$ such that $2^{\mu }=\tau$ for every cardinal $\mu$ with $\rho \le \mu <\kappa$, thenThus an eventual plateau of the power-set function below a singular cardinal continues at that cardinal. One can see the mechanism by decomposing $\kappa$ into $\mathrm{cf}(\kappa )$ bounded pieces: $2^{\kappa }=(2^{<\kappa }{)}^{\mathrm{cf}(\kappa )}$. On the plateau choose $\mu \ge \mathrm{cf}(\kappa )$, so ${\tau }^{\mathrm{cf}(\kappa )}=(2^{\mu }{)}^{\mathrm{cf}(\kappa )}=2^{\mu }=\tau$.

First the increasing sequence is cofinal in $\kappa$. Write $\mu ={\sup }_{\xi <\delta }{\kappa }_{\xi }$. Strict increase gives $|\delta |\le \mu$ when the sequence is infinite, so its cardinal sum is $\max (|\delta |,\mu )=\mu$. The given sum therefore implies $\mu =\kappa$ and $|\delta |\le \kappa$.

For any function $f:\lambda \to \kappa$, choose an index ${\xi }_t$ with $f(t)<{\kappa }_{{\xi }_t}$ for every $t<\lambda$. There are at most $\lambda <\mathrm{cf}(\delta )$ such indices, so they are bounded below $\delta$. Enlarging the bound if needed gives an index $\xi$ with $\mathrm{ran}(f)\subseteq {\kappa }_{\xi }$. ThusConversely each summand is at most ${\kappa }^{\lambda }$, and $|\delta |\le \kappa \le {\kappa }^{\lambda }$ because $\lambda \ne 0$. Infinite cardinal multiplication consequently givesCombining the inequalities proves ${\kappa }^{\lambda }={\sum }_{\xi <\delta }{\kappa }_{\xi }^{\lambda }$.

For the intended infinite regular cardinal $\kappa$, set $S_0(\delta )=0$. Suppose $0<\zeta <{\kappa }^{+}$ and the functions for its predecessors have been constructed. Since $|\zeta |\le \kappa$, choose a surjection $e_{\zeta }:\kappa \to \zeta$ and defineThe supremum is below $\kappa$: it uses fewer than $\kappa$ ordinals below $\kappa$, and $\kappa$ is regular. This defines a function $\kappa \to \kappa$ at each stage of the recursion.

For any $\alpha <\zeta$, choose $\eta$ with $e_{\zeta }(\eta )=\alpha$. Whenever $\delta >\eta$, the displayed supremum includes $S_{\alpha }(\delta )+1$, so $S_{\alpha }(\delta )<S_{\zeta }(\delta )$. Therefore the recursion yields a long chain under eventual domination of length ${\kappa }^{+}$.

Take the graphs $G_{\zeta }=\{(\delta ,S_{\zeta }(\delta )):\delta <\kappa \}$. Each has size $\kappa$, and any two meet in fewer than $\kappa$ points, because their functions are eventually strictly ordered. Fix a bijection $b:\kappa \times \kappa \to \kappa$ and let $A_{\zeta }=b[G_{\zeta }]\in [\kappa {]}^{\kappa }$.

Given $I\in [{\kappa }^{+}{]}^{\kappa }$, enumerate it as $\{{\zeta }_{\eta }:\eta <\kappa \}$ and trim the sets in this order:The deleted part is a union of fewer than $\kappa$ pairwise intersections, each of size less than $\kappa$. Regularity ensures its size is less than $\kappa$. If $\nu <\eta$, then $A_{{\zeta }_{\eta }}^{\ast }$ is disjoint even from the full earlier $A_{{\zeta }_{\nu }}$, hence from its trimmed version. Thus all the retained sets are pairwise disjoint and each lost fewer than $\kappa$ elements. This provesThe argument is the essential disjointness of small subfamilies of a regular-cardinal almost disjoint family.

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.

Start with a ground model containing a normal splitting Suslin tree $T$, and force with $\mathrm{Add}(\omega ,{\omega }_2)$, the finite partial functions from ${\omega }_2\times \omega$ to $2$. Such a starting model is relatively consistent with ZFC: the constructible universe has $◊$, and $◊$ implies the existence of a Suslin tree.

This Cohen forcing is Knaster. In an uncountable family of finite conditions, the delta-system lemma yields an uncountable family whose domains have one common root; thinning to common values on that finite root makes its members pairwise compatible. If $P$ is Knaster and $T$ is CCC, the product $P\times T$ is CCC: thin any uncountable family to pairwise compatible $P$ coordinates, then use CCC to find two compatible $T$ coordinates. Thus $P$ forces that the old $T$ is still CCC A name for an uncountable antichain would otherwise produce an uncountable antichain in this product by deciding its nodes.

Countable levels and the tree's splitting property persist, and the splitting argument from the previous part rules out a new cofinal branch in a CCC tree. Therefore the old $T$ is still Suslin. Meanwhile the forcing adds at least ${\aleph }_2$ distinct reals, with all cardinals preserved, so $2^{{\aleph }_0}\ge {\aleph }_2$.

We obtain a model with a Suslin tree and $\neg \mathrm{CH}$. Hence existence of a Suslin tree does not imply the Continuum hypothesis. This uses Knaster forcing preserves Suslin trees, not a claim that arbitrary CCC forcing preserves them.

Choose a name $\dot{f}\in M$ for $f$ and a condition $p\in G$ forcing that it is a function $\alpha \to \beta$. The ordinals $\alpha ,\beta$ belong to $M$, by forcing preserves ordinals. For each $\zeta <\alpha$, choose in $M$ a maximal forcing antichain $A_{\zeta }$ in the cone above $p$, every member deciding $\dot{f}(\zeta )$ as an ordinal below $\beta$. Conditions deciding an ordinal-valued name are dense, by the forcing theorem.

Let $y(\zeta )$ be the set of values decided by members of $A_{\zeta }$. The chain condition gives $|y(\zeta ){|}^M<\kappa$, and Choice and Replacement in $M$ assemble all these sets into a function $y\in M$ with domain $\alpha$. Since $p\in G$, genericity ensures that $G$ meets each deciding forcing antichain above $p$; equivalently enlarge it to a global maximal forcing antichain by conditions incompatible with $p$. Its chosen value is the actual $f(\zeta )$. ThereforeThis is the possible-values lemma for chain-condition forcing. All size bounds in its construction are internal to the ground model.

Suppose a ground cardinal $\lambda \ge \kappa$ became smaller in the extension. There would be a surjection $f:\alpha \to \lambda$ for some ordinal $\alpha <\lambda$. A finite domain cannot map onto an infinite ordinal, so use part (a) to cover $f$ by a ground function $y$ of small value sets. Its range is contained in $U={\bigcup }_{\zeta <\alpha }y(\zeta )$.

If $\lambda >\kappa$, the ground cardinality of $U$ is at most $|\alpha {|}^M\cdot \kappa <\lambda$. If $\lambda =\kappa$, regularity bounds a union of fewer than $\kappa$ sets of size less than $\kappa$ by a cardinal less than $\kappa$. In either case $U$ is a proper ground subset of $\lambda$, and the same missing ordinal remains missing in the extension, contradicting surjectivity.

Hence all ground cardinalities at least $\kappa$ are preserved. Ground bijections also remain bijections, so old noncardinals cannot turn into cardinals. The assertion does not prevent collapse of smaller cardinals, and consequently does not by itself preserve the aleph index of $\kappa$.

Work internally in the ground model and write $P=\mathrm{Levy}({\aleph }_0,\lambda )$. Conditions are finite partial functions with the coordinate-wise value bounds in the PDF, ordered by inclusion, so compatible conditions agree on their common domain and their union is a common strengthening.

Suppose there were $\lambda$ many pairwise incompatible conditions. Each fixed finite domain carries fewer than $\lambda$ possible functions: it uses finitely many ordinals below $\lambda$, and each coordinate has fewer than $\lambda$ choices. Regularity therefore lets us select $\lambda$ distinct domains. The finite-set form of the generalized delta-system lemma gives a subfamily of size $\lambda$ with common intersection $r$. There are fewer than $\lambda$ possible value assignments to the finite root $r$. Regularity gives a further subfamily of size $\lambda$ agreeing on the root. Any two of its conditions have a union in $P$, contradicting incompatibility. Thus $P$ has the $\lambda$-chain condition. Strong inaccessibility is more than is needed for this finite-support argument; regular uncountability suffices.

For each $0<\alpha <\lambda$ and $n<\omega$, requiring $(n,\alpha )$ to be in the domain is dense. For each $\xi <\alpha$, requiring $\xi$ to occur as a value at some $(n,\alpha )$ is also dense: choose a fresh $n$ and extend the condition. Hence the generic union defines, for each such $\alpha$, a surjection $g_{\alpha }:\omega \to \alpha$. Every ground ordinal below $\lambda$ is therefore countable in $M[G]$.

By the chain condition and the cardinal-preservation argument, $\lambda$ remains an uncountable cardinal in the extension; the same small-value argument preserves its regularity as well. All ordinals are unchanged. Since every smaller ordinal is countable and $\lambda$ is not,This is the finite Lévy collapse to omega-one. The dense-set construction proves the collapse below $\lambda$, while the chain condition is what prevents collapsing $\lambda$ itself.

For a smooth function, pullback of a smooth function is composition. The chain rule givesFor a vector field $Y$, use any local coordinates. To first order,Multiplying this inverse differential by $Y({\varphi }_s(x))$ givesThusThis is the Lie bracket of vector fields, whose action on a function is $X(Yf)-Y(Xf)$.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.

Away from the origin, differentiate $r^2=x_j{x}_j$ using Einstein notation:HenceThis gradient is the radial unit vector. Division by $r$ is legitimate on the punctured domain; the distance function is not differentiable at the origin.

The divergence formula reduces the problem to the ordinary differential equation $r{f}^{\prime }+nf=0$ on $r>0$. Multiplication by $r^{n-1}$ makes it an exact derivative:Since the interval $r>0$ is connected, $r^{n}f(r)$ is a single constant $C$. Conversely, substituting $f=C{r}^{-n}$ into the divergence formula gives zero. ThusThis proves both existence and uniqueness up to the constant within the specified radial class. For $n=1$, the two punctured half-lines still share this same constant because the stipulated coefficient is one function of $r=|x|$.

Away from the vertex, use the defining function $H=z^2-x^2-y^2$. Its gradient is normal to the smooth part of the double circular cone, andon the surface. Hence the field is tangent at every regular point of the cone. At the origin the cone has no unique tangent plane, but $\mathbf{F}(0)=0$, which belongs to its tangent cone. An equivalent argument, valid also at the vertex, uses the integral curves of a vector field: $\dot{\mathbf{x}}=\mathbf{x}$ has solution $\mathbf{x}(t)=e^t\mathbf{x}(0)$, and these dilations preserve $H=0$. This is an example of radial dilation tangency to homogeneous zero sets.

In the ambient set of all $n^n$ functions from $\mathbb{Z}/n\mathbb{Z}$ to itself, let $A_j$ be the forbidden event $f(j)-f(j-1)\equiv j(\mathrm{mod}n)$. Apply the avoidance form of the inclusion-exclusion principle from part (i).

For a proper subset $J$ of the $n$ cyclic difference constraints, with $|J|=k<n$, the selected edges form disjoint paths around the cycle, including isolated vertices. There are $n-k$ components. On each component, choosing one value of $f$ arbitrarily determines all the other values by the prescribed differences. No consistency condition occurs because the cycle has been broken. ThereforeThis also works for $n=2$, whose two directed constraints use the same pair of vertices: any single constraint leaves one freely chosen value.

If all $n$ constraints hold, summing them around the cycle gives the necessary conditionIt holds when $n$ is odd and fails when $n$ is even, the latter sum being congruent to $n/2$. When it holds, choose $f(0)$ in $n$ ways and determine every other value; the final constraint is then consistent. Thus the full intersection has size $F=n$ for odd $n$ and $F=0$ for even $n$.

There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ subsets of size $k$. By the binomial theorem,Hence the number of admissible functions is