Use only the variable names $x,y$. Define mutually renamed formulas bywhere $D_m^y$ is obtained from $D_m^x$ by globally interchanging $x$ and $y$, including their bound occurrences. By induction, $D_m^x(a)$ means that an increasing chain of $m$ elements starts at $a$. Henceholds exactly when the linear order has at least $n$ elements. Inner quantifiers may bind the name $x$ again, but their scope is only the nested subformula; the comparison $x<y$ still refers to the preceding point. This finite-variable logic construction works on finite and infinite linear orders.

The sheaf $A_n$ is noninitial: it contains $y_n$, whose value at $n$ includes $1_n$. In a two-valued topos its support, the image of $A_n\to 1$, can therefore only be $1$. An image equal to $0$ would force $A_n=0$, since the initial object of a topos is strict. Consequently every $A_n\to 1$ is an epimorphism.

Products of sheaves on a small site exist and are computed as products of the underlying presheaves, since a product of unique matching-family amalgamations again uniquely amalgamates. For the countable product $P={\prod }_{n\in \mathbb{N}}{A}_n$, its value at stage $m$ is ${\prod }_n{A}_n(m)$. The factor with $n=m+1$ is empty by the stronger statement proved in the preceding part. Thus $P(m)=\varnothing$ at every stage $m$, with the unique empty restriction maps. It is the empty presheaf, already an initial sheaf because no empty sieve on a category covers. HenceThis proves the final PDF-only request and exhibits the separating tower site with vanishing product. It is a genuine failure of countable choice in this Boolean two-valued sheaf topos; it does not contradict the classical external construction of the site.

First prove the hinted statement. Let $U\subseteq y_n$ be the domain of a partial presheaf map $\varphi :U\to y_{n+1}$. If $U$ is nonempty, take a section $f:m\to n$ in $U(m)$ and write $g={\varphi }_m(f):m\to n+1$. Hypothesis (c) supplies $h,k:m+1\to m$ with $fh=fk$ but $gh\ne gk$. Because $U$ is a subpresheaf, both restrictions belong to $U(m+1)$; naturality givesa contradiction. Therefore every such partial map has empty domain.

Now suppose a presheaf map $y_n\to A_{n+1}$ existed. Pull back the monomorphism $y_{n+1}\hookrightarrow A_{n+1}$ along it. Its unit image is dense by the preceding plus-construction argument, and density is stable under pullback, so the resulting domain $U\subseteq y_n$ is dense. It gives exactly a partial map $U\to y_{n+1}$ of the prohibited kind. Such a domain cannot be empty: at the section $1_n$ of $y_n(n)$, density would otherwise require the empty sieve on a category to cover. Thus there is no map $y_n\to A_{n+1}$.

Any sheaf map $A_n\to A_{n+1}$, composed with $y_n\hookrightarrow A_n$, would be such a presheaf map. HenceThe stronger intermediate statement also says $A_{n+1}(n)=\varnothing$, by the Yoneda correspondence. It will be useful for the last clause.

Let $y_n=C(-,n)$. Suppose two sections $f,g\in y_n(m)$ have equal restrictions on a covering sieve on a category $R$ on $m$. Such a sieve on a category is nonempty, so choose $h:l\to m$ in it. Then $fh=gh$. Applying hypothesis (b) with the two right-hand factors both $h$ gives $f=g$, since $f,g$ have the same source and target. Hence every representable is separated for $J$.

In the associated-sheaf construction, sections of $F^{+}$ are matching families modulo agreement after covering refinement. The map $F\to F^{+}$ identifies two original sections only if they agree on a covering sieve on a category. Thus it is monic when $F$ is separated. The plus construction is itself separated, and applying it a second time gives the associated sheaf; its second unit is monic as well. Applying this to $y_n$ proves the unit is a monomorphismThe same construction shows the unit image is dense: a section of $F^{++}$ is locally represented by sections of $F^{+}$, which are in turn locally represented by sections of $F$, and transitivity combines the two covering refinements.

We prove Booleanity directly from the dense-cover description. Let $B$ be a subsheaf of a sheaf $F$. Its pseudocomplement $N$ consists of sections $x\in F(c)$ such that no restriction along any arrow into $c$ belongs to $B$. This is a subpresheaf. It is also a sheaf: if a section of $F$ restricts into $N$ on a dense covering sieve on a category, any hypothetical restriction into $B$ has a further restriction in that cover; it would then belong to both $B$ and $N$, impossible. Thus the unique amalgamation in $F$ of an $N$-matching family lies in $N$.

For any $x\in F(c)$, the sieve on a category of arrows on which its restriction lies in $B$ or $N$ is dense. Given an arrow $u:d\to c$, either some further restriction belongs to $B$, supplying the required member of that sieve on a category, or none does, in which case the restriction along $u$ itself belongs to $N$. Therefore $B\cup N$ is locally all of $F$, so their join in the sheaf topos is $F$. Their intersection is empty. Every subsheaf consequently has a complement, proving the topos is Boolean.

For two-valuedness, let $U\hookrightarrow 1$ be a nonempty subterminal sheaf. At some stage $m$, its value is the singleton. For every $k\ge m$ there is an arrow $k\to m$, so restriction makes $U(k)$ the singleton as well. At any stage $n$, the sieve on a category of arrows into $n$ whose domains are at least $m$ is dense: after an arrow with domain $l$, choose a further domain $k\ge \max (l,m)$ and use the existence of arrows downwards. On this covering sieve on a category the unique terminal section has a matching family in $U$, so the sheaf condition gives $U(n)=1$. Hence $U=1$. Since the empty sieve on a category never covers, the empty presheaf is an initial sheaf distinct from $1$. Thusand the topos is two-valued, in the distinct global sense of a two-valued topos.

A dense sieve, also called a persistently nonempty sieve, is one whose every pullback sieve on a category is nonempty: a member factoring through $f$ has the form $fh$, corresponding to a member $h$ of $f^{\ast }R$. The maximal sieve on a category clearly has this property. If $R$ is persistently nonempty, then $u^{\ast }R$ is too: for any $v$ into the domain of $u$, apply density to $uv$ to find an extension $uvh\in R$, giving $vh\in u^{\ast }R$.

For transitivity, suppose $R$ is dense on $n$ and $S$ is a sieve on a category such that $r^{\ast }S$ is dense for each $r\in R$. Given $f:m\to n$, choose $h$ with $fh\in R$. Density of $(fh{)}^{\ast }S$, tested on its identity arrow, supplies $k$ with $fhk\in S$. Thus $S$ is dense. These are the maximality, pullback and transitivity axioms of a Grothendieck topology. Therefore the specified sieves on a category form the coverage $J$, namely the double-negation topology. The empty sieve on a category is never covering, because its identity pullback is empty.

For the Category of sets, take the monoidal tensor product to be the Cartesian product, $A\otimes B=A\times B$, with unit a singleton $1$. On morphisms use $(f\times g)(a,b)=(f(a),g(b))$. Define the associator by $((a,b),c)\mapsto (a,(b,c))$ and the unitors by $(\ast ,a)\mapsto a$ and $(a,\ast )\mapsto a$. These are natural isomorphisms, with inverses obtained by rebracketing or adjoining $\ast$. Both paths in the pentagon send $(((a,b),c),d)$ to $(a,(b,(c,d)))$, and both paths in the triangle send $((a,\ast ),b)$ to $(a,b)$. This proves the axioms elementwise.

There is a second structure: use disjoint union $A⨿B$ and unit $\varnothing$. Define the associator by preserving each element and its summand while changing the nested tags, and define the unitors by removing the empty summand. These are natural and invertible. In the pentagon, either path takes an element in any of the four summands to that same element in the same summand of $A⨿(B⨿(C⨿D))$; the triangle removes an empty intermediate summand on either path. Thus both axioms hold.

These product and coproduct monoidal structures on sets have different units, $1$ and $\varnothing$, which are not isomorphic. They are even inequivalent as monoidal categories: an equivalence of the underlying Category of sets preserves terminal objects, so it cannot take the singleton unit of the product structure to the empty unit of the coproduct structure. Hence the monoidal structure on sets is not unique.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a monoidal unit object $I$, and natural isomorphismsThe associator obeys the pentagon identity for a monoidal category,and the unitors obey the triangle identity for a monoidal category,Composition in a category here acts from right to left. The first equation compares two paths from $((A\otimes B)\otimes C)\otimes D$ to $A\otimes (B\otimes (C\otimes D))$; the second compares two ways of removing the intervening unit.

It is precisely a bicategory with one object $\ast$. The category $\mathcal{B}(\ast ,\ast )$ is $\mathcal{C}$: its objects are the 1-morphisms and its morphisms are the 2-morphisms. Horizontal composition in a category is $\otimes$, the identity 1-morphism is $I$, and the associator and unitors give the weak associativity and unit constraints. Vertical composition in a category is ordinary composition in a category in $\mathcal{C}$; the interchange law follows because $\otimes$ is a bifunctor. The bicategory pentagon and triangle axioms are exactly the two equations above. Conversely the sole hom-category of any one-object bicategory gives these data, with the convention that horizontal composition in a category uses the indicated ordering of tensor factors.

Write the adjunction unit and adjunction counit of $F⊣G$ as $\eta ,\epsilon$. Its induced monad is $T=GF$, with multiplication $\mu =G{\epsilon }_F$. The Eilenberg-Moore comparison functor isThe unit law for $a_d$ is $G{\epsilon }_d{\eta }_{Gd}=1_{Gd}$, a triangle identity for an adjunction. Naturality of $\epsilon$ at ${\epsilon }_d$ gives ${\epsilon }_{d}FG{\epsilon }_d={\epsilon }_d{\epsilon }_{FGd}$; applying $G$ gives $a_{d}T{a}_d=a_d{\mu }_{Gd}$. Naturality at $h$ gives the monad algebra morphism equation for $Gh$. Thus $K$ is well-defined.

For each $d$, consider the fork in $\mathcal{D}$It commutes by the naturality equation just used. Its image under $G$ is the fork $T^2(Gd)\rightrightarrows T(Gd)\to Gd$ of the preceding question, with parallel morphisms ${\mu }_{Gd},T{a}_d$ and final map $a_d$. It is split using $s={\eta }_{Gd}$ and $t={\eta }_{T(Gd)}$. By the assumed reflection, ${\epsilon }_d$ is the coequalizer of this pair in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ for $h,k:d\to e$, naturality gives $h{\epsilon }_d={\epsilon }_{e}FGh={\epsilon }_{e}FGk=k{\epsilon }_d$. A coequalizer is an epimorphism, so $h=k$.

For fullness, let $u:K(d)\to K(e)$ be a monad algebra morphism. Thus $u:Gd\to Ge$ satisfies $u{a}_d=a_{e}Tu$. Set $v={\epsilon }_{e}Fu:FGd\to e$. We verify explicitly that $v$ equalizes the two morphisms of the displayed fork. The monad algebra morphism equation givesNaturality of $\epsilon$ first at $Fu$ and then at ${\epsilon }_e$ givesThe coequalizer therefore supplies a unique $h:d\to e$ with $h{\epsilon }_d=v$. Applying $G$ yields $Gh{a}_d=Gv=a_{e}Tu=u{a}_d$. The map $a_d$ is split epic with section ${\eta }_{Gd}$, so cancellation gives $Gh=u$. This proves fullness, and the earlier cancellation proves faithfulness. HenceNo claim that $K$ is essentially surjective is needed; that stronger conclusion needs the corresponding existence hypotheses.

For $G:\mathcal{D}\to \mathcal{C}$, a functor-split coequalizer pair is a parallel pair $f,g:Y\rightrightarrows Z$ such that its image admits a split coequalizer in $\mathcal{C}$. In one orientation this means morphisms $q:GZ\to Q$, $s:Q\to GZ$ and $t:GZ\to GY$ satisfyThese equations prove that $q$ is a coequalizer: if $hGf=hGg$, then $h=hGft=hGgt=hsq$, so $hs$ factors $h$; uniqueness follows from $qs=1$. They remain true under every functor, making a split coequalizer an absolute colimit.

To say that $G$ reflects such coequalizers means that any fork $Y\rightrightarrows Z\stackrel{p}{\to }W$ in $\mathcal{D}$ whose image is a split coequalizer is itself a coequalizer in $\mathcal{D}$. This is a reflection assertion about an existing fork; it does not by itself require lifting every split fork from $\mathcal{C}$, or creating its splitting maps in $\mathcal{D}$.

For $(A,a)$, consider the reflexive free-algebra presentation of a monad algebraBoth parallel morphisms are monad algebra morphisms: ${\mu }_A$ satisfies the monad associativity law, while $Ta$ is the image under $F^T$ of $a$. The map $a$ is a monad algebra morphism because $a{\mu }_A=aTa$, and the same equation says it equalizes the pair.

Suppose $h:F^T(A)\to (B,b)$ is a monad algebra morphism with $h{\mu }_A=hTa$. Put $k=h{\eta }_A:A\to B$. Naturality of $\eta$ and the unit law giveMoreover, since $h$ is a monad algebra morphism,Thus $k$ is a monad algebra morphism and factors $h$ through $a$. If $k^{\prime }a=h$, then $k^{\prime }=k^{\prime }a{\eta }_A=h{\eta }_A=k$, proving uniqueness. Therefore every monad algebra is the stated coequalizer of free algebras.

The underlying fork is a split coequalizer: take $q=a$, $s={\eta }_A$ and $t={\eta }_{TA}$. Then $qs=1_A$, ${\mu }_{A}t=1_{TA}$, and $(Ta)t=sa$. Its splitting maps need not be monad algebra morphisms; the preceding argument separately proves the coequalizer in ${\mathcal{C}}^T$.

Define the free algebra functor $F^T$ by $F^{T}X=(TX,{\mu }_X)$ and $F^{T}f=Tf$. The monad laws make ${\mu }_X$ an algebra action, and naturality of $\mu$ makes $Tf$ a monad algebra morphism. For an algebra for a monad $(A,a)$, definewith proposed inverse $g\mapsto aTg$. This inverse is a monad algebra morphism, sinceOne composite is $aTg{\eta }_X=a{\eta }_{A}g=g$. For a monad algebra morphism $h$, the other is $aT(h{\eta }_X)=aThT{\eta }_X=h{\mu }_{X}T{\eta }_X=h$. Both formulas commute with composition in a category in $X$ and with monad algebra morphisms in $(A,a)$, so the bijection is natural. Thus

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying, at every object $A$,These are the associativity and unit laws. An algebra for a monad is $(A,a:TA\to A)$ withA monad algebra morphism $h:(A,a)\to (B,b)$ is a morphism $h:A\to B$ satisfying $ha=bTh$. Identity morphisms satisfy this equation, and if $h,k$ satisfy it, then $kha=kbTh=cTkTh=cT(kh)$. Hence these objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$, with the forgetful functor $U(A,a)=A$ and $Uh=h$.

Define $\mathcal{T}$ to be the family of all unions of members of $\mathcal{B}$, including the empty union. We will verify that this is a topology and that $\mathcal{B}$ is a basis of a topology. The two conditions play different roles: the covering condition supplies the whole space, while the local intersection condition supplies closure under finite intersections.

Use the least-upper-bound property as the least upper bound axiom: every nonempty set of real numbers bounded above has a least upper bound in $\mathbb{R}$. It implies convergence of a bounded increasing sequence: if $L$ is the supremum of its terms, then for every $\epsilon >0$ some term exceeds $L-\epsilon$, and all subsequent terms lie between that term and $L$. A bounded decreasing sequence converges by applying this argument to its negative.

The alternating series test states that if $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n-1}{b}_n$ converges. To prove it, let $S_N$ denote its partial sums. The even partial sums are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge 0$. The odd partial sums are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le 0$. Moreover,Completeness of the real numbers therefore gives a limit of a sequence for each subsequence, denoted $L_e,L_o$ respectively for the even and odd subsequences. Their difference is $S_{2m+1}-S_{2m}=b_{2m+1}\to 0$, so $L_e=L_o$. Every sufficiently late partial sum belongs to one of these two subsequences, proving convergence of the full sequence of partial sums. Multiplying the terms by $-1$ leaves convergence unchanged.

Let $f$ be continuous on a closed bounded interval $[a,b]$. First prove that $f$ is a bounded function. If it failed, we could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x_{\ast }$. Because the interval is closed, $x_{\ast }\in [a,b]$. Continuity gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j\to \infty$.

Now let $M=\sup \{f(x):x\in [a,b]\}$, which exists by completeness of the real numbers and the bound just proved. Choose $y_n\in [a,b]$ with $f(y_n)>M-1/n$. A convergent subsequence has limit of a sequence $y_{\ast }\in [a,b]$, and continuity gives $f(y_{\ast })=M$. Applying the same argument to $-f$ gives attainment of the infimum. This proves the extreme value theorem: the maximum and minimum both exist and are attained. The argument also covers a one-point interval, where the conclusions are immediate.

Let $N$ be the $t\times b$ incidence matrix of a set system. The balanced incomplete block design overlap counts give $N{N}^T=(r-\lambda )I+\lambda J$. On the zero-sum subspace its eigenvalue is $r-\lambda =r(t-k)/(t-1)>0$; on the constant vector its eigenvalue is $r+(t-1)\lambda >0$. Therefore $N{N}^T$ has matrix rank $t$. Since a product cannot have larger rank than either factor,This proves Fisher's inequality for block designs directly. The incomplete, nontrivial hypotheses ensure the positive eigenvalue needed for this argument.

The identities require $r=k$ and $k(k-1)=2\cdot 3=6$, so $k=r=3$. Take treatments $1,2,3,4$ and the four blocksEach treatment is absent from exactly one block, hence occurs three times. Each pair is absent from the two blocks missing one of its members, hence occurs together in the other two. This is the required symmetric balanced incomplete block design; here $r-\lambda =1$ is indeed a square.

Symmetry forces $r=k$, and the pair-counting identity becomes $k(k-1)=2\cdot 45=90$. Its only positive integral solution is $k=r=10$. Since the number of blocks is even, the even-order symmetric design square obstruction requires $r-\lambda =8$ to be a square. It is not. No design with these parameters exists.

The transformed values are Cox–Snell residuals. Under a correct common survival function and a good fitted cumulative hazard function, the complete transformed failure times should have unit exponential distribution. Transform the observed times and retain their original censoring indicators, then estimate residual survival with a Kaplan–Meier estimator. Its target isso a plot of its logarithm against $u$ should lie approximately on a line of slope $-1$. Censored residuals must not be treated as fully observed failure times.

There is a useful caution for this particular common nonparametric fit. The shared-fit Cox–Snell residuals reproduce the fitted survival curve property means that the pooled residual plot is largely built into the transformation. With event ordering and compatible tie processing preserved, at a transformed failure time $u_j=\hat{H}(t_j)$ its residual Kaplan–Meier estimator equals $\hat{S}(t_j)=e^{-u_j}$. Finite steps, coalesced censoring times and estimation uncertainty still matter, and an infinite final residual arises if the fitted survivor becomes zero. Separate group plots are more informative about a common-distribution assumption than this pooled fit alone.