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A monotone map preserving suprema of nonempty directed sets. Between pointed domains it need not send bottom to bottom. The pointwise directed supremum of a directed family of such maps is again Scott continuous: interchange the two directed suprema in ${\sup }_f{\sup }_{x\in A}f(x)$.

Take two increasing countable chains sharing a least element $\perp$ and a greatest element $\top$, with interiors $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$ and no comparisons between branches. Each entire interior chain has $\top$ as its least upper bound. A directed set meeting both interiors must contain $\top$. Otherwise it lies within one branch, where it has either a greatest member or least upper bound $\top$. This proves that the partially ordered set is a directed-complete partial order.

The only element way-below $a_n$ is $\perp$. Indeed, an element way-below $a_n$ must be below $a_n$, but every positive $a_m\le a_n$ fails the way-below relation when tested against $\{b_1,b_2,\dots \}$: its least upper bound is $\top \ge a_n$, yet it has no member above $a_m$. Nonempty directed sets always contain an element above $\perp$, so $\perp \ll a_n$. Thus the elements way-below $a_n$ have least upper bound $\perp \ne a_n$, and this is not a continuous poset.

Two projective triangles are in perspectivity from a point when the three projective lines joining corresponding vertices meet in one projective point, called their center of perspectivity.

Two projective triangles are in perspectivity from a line when the three intersections of corresponding sides lie on one projective line, called their axis of perspectivity. In the nondegenerate setting this is dual to perspectivity from a point under point-line duality.

Every element of a pointwise definable model is uniquely specified by a formula with no added parameters. Finite tuples are then uniquely specified by conjunctions of their defining formulas. For a complete theory these formulas isolate the tuple types, so such a model is an atomic model. The converse fails: a countable dense linear order without endpoints with a named point is atomic, while its other points are moved by structure automorphisms fixing that point.

An injective self-map of the complex projective line preserving all cross-ratios is a projective transformation; neither continuity nor prior surjectivity is needed. Choose the unique projectivity agreeing with the map on three distinct projective points and compose with its inverse. The resulting map fixes those three points. Normalize them to $\infty ,0,1$. Preservation of the cross-ratio $(\infty 0:1z)=z$ now fixes every other point as well, proving that the normalized map is the identity map. Undoing the composition proves the assertion.

For the two-dimensional heat equation with zero Dirichlet boundary conditions, the Crank-Nicolson diffusion scheme preserves the interval between the minimum and maximum of the initial nodal values and zero under the displayed restriction. The old-time coefficients form a convex combination, while the new-time matrix has positive diagonal, nonpositive off-diagonal entries and row diagonal-dominance margin one. A hypothetical new value above the old maximum makes the new-time row at least that value while its right side is at most the old maximum, a contradiction; apply the reversed inequality at a new minimum. In particular initial nonnegative or nonpositive values preserve their sign. The step restriction guarantees these value bounds even though scalar A-stability holds without it.

Give the real interval $[0,1]$ its usual total order. Every nonempty directed set has a least upper bound in the real interval, so it is a directed-complete partial order. If $x<y\le \sup D$, some $d\in D$ exceeds $x$; otherwise $x$ would be an upper bound for $D$. Also $0\ll 0$, since every member of a nonempty $D$ is at least zero. For $x>0$, the increasing sequence $x(1-1/n)$, $n\ge 1$, has least upper bound $x$ but never reaches it, so $x/\ll x$. Together with $x\ll y\Rightarrow x\le y$, this proves the displayed way-below relation.

For each $y>0$, the elements way-below $y$ form $[0,y)$, a directed set with least upper bound $y$; for $y=0$ they form $\{0\}$. Thus this is a continuous poset. Its only compact element is zero, so it is not an algebraic partially ordered set.

If the ambient power set of $\omega$ is not well-orderable and $a\in L$, then the full ambient power set of $a$ belongs to $L$ exactly when $a$ is finite. Finite set operations give the forward construction for finite $a$. If $a$ is infinite, its constructible well-order gives an injection $\omega \to a$. The induced injection of power sets would make the ambient power set of $\omega$ well-orderable if $\mathcal{P}(a)$ belonged to $L$.

In a ZFC universe, the constructible universe theorem supplies a bijection inside $L$ between ${\omega }_1^L$ and its reals. The same graph is a bijection externally. Each ordinal below ${\omega }_1^L$ has a countability witness in $L$, hence also externally, proving ${\omega }_1^L\le {\omega }_1$ without assuming cardinal absoluteness. Therefore the constructible reals have ambient cardinality at most ${\aleph }_1$. If the ambient continuum is larger, some subset of $\omega$ is not constructible. The ordinal ${\omega }_1^L$ may be externally countable, so the argument is an upper bound and not a claim of equality.

For planar Brownian motion started at $p+iq$, $p>0$ and $0<q<a$, in $\{x+iy:x>0,0<y<a\}$, the displayed value is its probability of exiting through $y=a$. The half-strip conformal map by hyperbolic cosine sends the top edge to $(-\infty ,-1)$ and the starting point to $A+iB$, where $A=\cosh (\pi p/a)\cos (\pi q/a)$ and $B=\sinh (\pi p/a)\sin (\pi q/a)$. Integrating the Cauchy exit law from a Brownian half-plane over that ray proves the formula. As $p\to \infty$ it tends to $q/a$.

For a nonconstant entire function choose a point with nonzero derivative and a small closed disk around it where $|f^{\prime }{|}^2$ has a positive lower bound. recurrence of planar Brownian motion gives infinitely many successive crossings from a concentric inner circle to the disk boundary. Their durations are positive iid variables by strong Markov and rotational invariance of Brownian motion. The strong law applied to their bounded positive truncations makes their sum infinite. Hence the conformal clock is infinite and the analytic Brownian image, under time change of a continuous process, has full infinite Brownian lifetime. Centering the disk at a noncritical point is essential.

On the degree-$n$ homogeneous polynomials use $e_k=x^{n-k}{y}^k$. Then $P{e}_k=k{e}_{k-1}$ and $Q{e}_k=(n-k)e_{k+1}$; both maps are nilpotent of index $n+1$. Their commutator is diagonal in this basis with distinct eigenvalues $n,n-2,\dots ,-n$. Thus the raising and lowering operators are not diagonalizable for $n\ge 1$, while their commutator is.

Let $S$ be a clamped cubic spline interpolant and let $T$ have the same values at all spline knots and the same endpoint slopes. For $T\in C^2[x_0,x_n]$, or more generally $T^{\prime }$ an absolutely continuous function with square-integrable second derivative, the exact displayed identity holds. In particular $S$ uniquely minimizes the second derivative roughness penalty $J$ among these competitors.

To prove the identity, put $E=T-S$. On each knot interval, two applications of integration by parts give $\int S^{\prime \prime }{E}^{\prime \prime }=[S^{\prime \prime }{E}^{\prime }-S^{\prime \prime \prime }E]$, since the cubic piece has zero fourth derivative. Terms containing $E$ vanish at the spline knots; terms containing $E^{\prime }$ cancel at interior spline knots because $S^{\prime \prime }$ and $E^{\prime }$ are continuous. The endpoint terms vanish because $E^{\prime }$ is zero there. Thus $\int S^{\prime \prime }{E}^{\prime \prime }=0$, and expanding $(S^{\prime \prime }+E^{\prime \prime }{)}^2$ proves the identity. Equality makes $E^{\prime \prime }=0$, so $E$ is an affine function; a prescribed endpoint value and slope force $E=0$.

If $T$ also matches the first derivative of $S$ at every spline knot, both $E$ and $E^{\prime }$ vanish at the ends of each knot interval. For a sufficiently differentiable $T$, integration by parts separately on that interval gives the stronger local identity $\int E{T}^{(4)}=\int (E^{\prime \prime }{)}^2\ge 0$.

An adjustable-cistern barometer associated with Jean Nicolas Fortin. Restoring the cistern surface to a fixed index makes the lower mercury level reproducible before the upper level is read.^[ref]

At its choice time this claim permits purchase of either of two later-expiring European options for an exercise cost $K$, or abandonment. Its payoff is $\max (p-K,c-K,0)$. For common strike $K_3$ and remaining time $\tau$, put-call parity places the branch crossing at $D=K_3{e}^{-r\tau }$. In the Black-Scholes model, both branch values there equal $D[2\Phi (\sigma \sqrt{\tau }/2)-1]$. An exercise cost above this minimum creates a central interval in which neither choice is worth exercising.

For a real symmetric positive-definite matrix, the lower triangular matrix $C$ with positive diagonal in its Cholesky decomposition is its lower Cholesky factor; $C^T$ is the corresponding upper factor. If a full-column-rank matrix $B$ has a reduced QR decomposition $B=QR$ with positive diagonal of $R$, then $B^{T}B=R^{T}R$, so $R$ is the upper Cholesky factor of $B^{T}B$.

A hypothetical finite chief series in $\mathbb{Z}$ has a first nonzero term $n\mathbb{Z}$. The subgroup $2n\mathbb{Z}$ is normal in $\mathbb{Z}$ and strictly between zero and that term. This contradicts the defining maximality condition for the first chief factor. Finite groups always admit chief series, by choosing a maximal chain of their normal subgroups; this example shows why finiteness matters.

For initial data $u(x,0)=1+x^3/4$ on $0\le x\le 1$, the characteristic surface has $x=s{e}^r$, $y=3s^2(1-e^{-2r})/2$, $u=1+s^3(e^{-3r}-3e^{-r}/4)$. Its projection determinant is $3s^2{e}^r(2e^{-2r}-1)$, so the sheet connected to the initial data loses local invertibility at $r=\log 2/2$. The regular sheet has $r<\log 2/2$ and projects to $0<x<\sqrt{2}$, $3(1-x^{-2})/2<y<3x^2/8$. Its upper boundary is a fold point of a critical manifold caustic. Solving the inverse yields $s^2=[x^2+\sqrt{x^4-(8/3)x^{2}y}]/2$; the other square-root branch lies on the folded sheet and cannot simply be joined as a unique smooth solution. The initial point $s=0$ is already characteristic, illustrating the need to inspect endpoints as well as the interior of a Cauchy curve.