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For a complex potential $w(z)=m/(z-a)$, reflection in the straight wall through zero at angle $\theta$ is $Rz=e^{2i\theta }\bar{z}$. The image construction $W(z)=w(z)+\overline{w(Rz)}$ givesOn the wall $Rz=z$, this is real, so its stream function is constant there. This proves the impermeability condition. Thus the image position is $a^{\prime }=e^{2i\theta }\bar{a}$ and its moment is $m^{\prime }=e^{2i\theta }\bar{m}$. If the point dipole is advected by the regular image flow, the complex velocity convention $W^{\prime }=u-iv$ giveswhere $d=\mathrm{Im}(e^{-i\theta }a)$ is its signed distance from the wall. The last equality follows from $a-a^{\prime }=2i{e}^{i\theta }d$. It applies when $d\ne 0$.

A planar potential dipole at $a$ with complex moment $m$ has complex potential of a two-dimensional potential flow $w(z)=m/(z-a)$. Its complex velocity is $w^{\prime }(z)=-m/(z-a{)}^2$, interpreted as $v_x-i{v}_y$. Unlike the three-dimensional potential dipole, this is a two-dimensional singularity of the logarithmic-source potential. For example, two opposite source coefficients $s$ at $a\pm \delta /2$ have potential $s[\log (z-a-\delta /2)-\log (z-a+\delta /2)]$, whose leading term is $-s\delta /(z-a)$ as the separation tends to zero with $s\delta$ fixed. This gives the dipole as a limiting source–sink pair, with the complex sign and orientation included in $m$.

A nonreal complex number is a complex number with nonzero imaginary part. Being nonreal is a property of one number; forming a complex conjugate pair additionally specifies the relation between two numbers.

The complete theory of the standard successor structure is axiomatized by injectivity of $s$, $0$ outside its range, a predecessor for every nonzero element, and the sentences $s^n(x)\ne x$ for each positive integer $n$. Its models have a single distinguished one-sided chain generated by zero and any collection of two-sided integer chains. The classification of models of the successor theory proves completeness and explains why the theory is categorical at every uncountable cardinal but not at the countable cardinal.

In the first-order language of logical equality, assert that there are at least $n$ elements for every positive integer $n$. Eliminating one existential logical variable either substitutes a logical equality or avoids finitely many points, so this first-order theory has quantifier elimination and is complete. It is categorical in every infinite cardinal. Its complete $n$-types are the logical equality partitions of $n$ coordinates and are counted by the Bell numbers.

The first-order theory of a bijection $f$ with named inverse $g$ and no finite cycles has first-order models consisting of disjoint integer orbits. Atomic logical formulas reduce to $x_j=f^k(x_i)$ for integer $k$. An existential logical variable is either fixed by such a logical equality or must avoid finitely many points. This proves quantifier elimination and completeness. There is one complete one-type; the complete two-types are the integer offsets and the nonprincipal different-orbit partial type. There are countably many finite-arity partial types but countable first-order models with different numbers of orbits are not isomorphic.

A topological space is compact exactly when every family of closed sets with the finite intersection property has nonempty intersection. Take complements and apply De Morgan's laws: an empty total intersection corresponds to an open cover, while an empty finite intersection corresponds to a finite subcover. This works without any Hausdorff condition. With the convention that an empty-family intersection is the whole ambient set, the finite intersection property includes nonemptiness of that ambient set, and the criterion correctly treats the empty compact space.

Every closed subset $F$ of a compact space $X$ is compact in its subspace topology. An open cover of $F$ by ambient open sets, together with $X\setminus F$, covers $X$. A finite subcover for $X$ still covers $F$ after the extra member is discarded. The result does not require $X$ to be Hausdorff.

Let $X$ be a Hausdorff space, let $K_n$ be a nested sequence of nonempty compact subsets of $X$, and let $f:X\to Y$ be a continuous map into a T1 space. Then the displayed equality holds. The forward inclusion is true for every map. For the reverse, fix $y\in {\bigcap }_{n}f(K_n)$. Each $F_n=K_n\cap f^{-1}(\{y\})$ is nonempty and closed in the compact set $K_1$: the $K_n$ are closed by the Hausdorff assumption, while the fiber is closed because $Y$ is T1 and $f$ is continuous. The nested sets $F_n$ have the finite intersection property, so compactness gives a point in their intersection. Its image is $y$.

The proof only needs the relevant fibers to be closed in $K_1$, rather than all of $Y$ to be T1. A Hausdorff space as codomain is a familiar sufficient special case. Mere continuity into an arbitrary space does not suffice.

If $K$ and $L$ are compact subsets of two spaces and $W$ is an open neighborhood of their product, there are open sets $U,V$ with $K\times L\subseteq U\times V\subseteq W$. First use compactness of $L$ to produce a rectangle around each slice $\{x\}\times L$, intersecting finitely many first factors and uniting second factors. Then use compactness of $K$, uniting finitely many resulting first factors and intersecting second factors. No Hausdorff assumption is required.

For $1<p<\infty$, let $q=p/(p-1)$ and partition the coordinates into consecutive finite blocks $B_k$ of lengths $m_k\to \infty$. The displayed map defines a norm-one operator on the l-p sequence space. By Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$, giving the upper norm bound. For the unit vector $e_j$ with $j\in B_k$, $T{e}_j=m_k^{-1/q}{e}_k$, so its norm tends to zero as $j\to \infty$.

Nevertheless, $x^{(k)}=m_k^{-1/p}{1}_{B_k}$ has norm one and maps to $e_k$. Those images are pairwise distance $2^{1/p}$ apart. Thus $T$ is not a compact operator, and has norm exactly one. Vanishing images of a chosen basis do not alone test compactness on ${\ell }^p$ for $p>1$. The map is the adjoint of the isometric block embedding $(Uy{)}_j=m_k^{-1/q}{y}_k$, $j\in B_k$, on ${\ell }^q$.

Suppose a linear map $S:c_0\to Y$ has compact image of the closed unit ball. It is bounded, and each $g\in Y^{\ast }$ attains its maximum on this compact image. Thus $S^{\ast }g$ attains its norm, so norm attainment on the vanishing sequence space implies $S^{\ast }(Y^{\ast })\subseteq c_{00}$. The Baire category theorem applied to the inverse images of the first $n$ coordinate spans makes this adjoint finite rank: these are closed linear subspaces covering the complete continuous dual space $Y^{\ast }$, so one has interior and is all of $Y^{\ast }$. Finally the Hahn-Banach theorem identifies $\ker S$ as the common kernel of $S^{\ast }(Y^{\ast })$, proving finite rank of $S$. Ordinary compact operators require only relative compactness of the ball image, so they need not have finite rank.

For a bounded linear operator into a Banach space, compactness is equivalent to the displayed condition, where ${\pi }_Y$ is the quotient map by a finite-dimensional subspace of the target. A finite net for the image of the unit ball gives such a subspace by taking its span. Conversely, if the image lies within $\epsilon$ of a finite-dimensional subspace, choose approximants there; they remain bounded because the original image is bounded. A finite net for these approximants gives a slightly larger finite net for the image. This proves total boundedness and hence compactness. This condition does not require choosing the approximants by a linear map, and therefore does not assume an approximation property for the Banach space.

The operator in the title is bounded and compact on the absolutely summable sequence space. The norm of a matrix operator on ${\ell }^1$ is the supremum of its absolute column sums: the triangle inequality gives one bound and coordinate unit vectors give the other. Here the column sum is $c_m={\sum }_{n=1}^{m-1}n!/m!\le 2/m$, using ${\sum }_{n=1}^{k}n!\le 2k!$. The latter inequality follows by induction. Retaining the first $N$ columns gives a finite-rank operator and a remainder of norm at most $2/(N+1)$. A norm limit of finite-rank operators is a compact operator, proving the assertion. This column-decay method can prove compactness even when each row has infinitely many nonzero entries.

For a vector field $V$, the tensorial second derivative is ${\nabla }^{2}V(X,Y)={\nabla }_X{\nabla }_{Y}V-{\nabla }_{{\nabla }_{X}Y}V$. With $R(X,Y)V={\nabla }_X{\nabla }_{Y}V-{\nabla }_Y{\nabla }_{X}V-{\nabla }_{[X,Y]}V$ and geometric torsion tensor $\mathcal{T}$, antisymmetrization givesThis follows by subtracting the two definitions and using ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]+\mathcal{T}(X,Y)$. By contrast, the directional operator commutator subtracts no derivative-index connection and obeys $[{\nabla }_X,{\nabla }_Y]V=R(X,Y)V+{\nabla }_{[X,Y]}V$. On scalars the directional commutator is $[X,Y]f$, while the tensorial Hessian commutator is $-\mathcal{T}(X,Y)f$. Distinguishing these objects prevents an apparent contradiction when a scalar Hessian has an antisymmetric part despite commuting coordinate vector fields.

An undoable command records enough information to restore its input state when applied immediately after its corresponding edit. In a sequence, insertion at index $i$ is reversed by deletion at $i$; deletion is reversed by reinsertion of the saved value. If a move places an entry at final index $j$ after removing it from index $i$, the inverse moves it from $j$ back to $i$. Removing the moved entry recovers the original sequence with exactly that entry missing, which proves the inverse for both directions and for $i=j$.

These local inverse identities extend to a history by applying inverses in reverse chronological order. A stack of completed commands and a stack of undone commands maintain that order. A fresh command invalidates the old redo branch. Arbitrary inverses after unrelated intervening edits need not satisfy the restoration identity. State snapshots in the Memento pattern supply a different way to realize the same chronological undo/redo contract.

For an integer $p\ge 2$ and $k\ge 0$, unlimited denominations $1,p,\dots ,p^k$ admit optimal greedy algorithm. In a minimum-count solution, no smaller denomination can occur $p$ times: carrying those stamps to the next denomination reduces their count by $p-1$. Thus all counts below the maximum are base-$p$ digits, whose total value is at most $p^k-1$. The maximum-denomination count is forced to be the quotient on division by $p^k$, and repeating the argument forces every greedy remainder step.

For a set $X$ with its cofinite topology, a subset $S$ is a connected subset exactly when $S$ is empty, a singleton, or infinite. Indeed, its subspace topology is the cofinite topology on $S$: the traces of open sets are precisely $S\setminus F$ for finite $F\subseteq S$, together with the empty set. For infinite $S$, two nonempty open sets intersect, because their complements have finite union. Hence $S$ is a hyperconnected space and therefore a connected space. A finite $S$ has the discrete topology, which admits a separation of a topological space as soon as $S$ has at least two points.

Let $Z$ be the underlying set of real numbers with the cofinite topology. The diagonal $W=\{(t,t):t\in \mathbb{R}\}$ in $Z\times \mathbb{R}$ with its subspace topology is homeomorphic to the usual real line. Indeed, $t\mapsto (t,t)$ is a continuous bijection: its first coordinate is continuous because usual finite sets are closed, and its second coordinate is the identity. The inverse is the second coordinate projection restricted to $W$. Consequently $W$ is not compact, whereas $Z$ is compact by the fact that every subset of a cofinite space is compact. Thus $W$ is not homeomorphic to $Z$.

For a map $A:M\to X^{\ast }$ on a subset of a real Banach space, coercivity with respect to $u_0\in M$ means the displayed limit as $u\in M$ and $\Vert u-u_0\Vert \to \infty$. Here the brackets denote the dual pairing. This growth condition is useful in existence theorems for monotone operators. A linear Hilbert-space estimate $⟨Au,u⟩\ge c\Vert u{\Vert }^2$, $c>0$, implies the condition at zero. Monotonicity alone does not: the zero operator is monotone and fails this growth condition on an unbounded space.