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For a symmetric interval of the form $[L,U]=[\hat{\theta }-c\mathrm{SE}(\hat{\theta }),\hat{\theta }+c\mathrm{SE}(\hat{\theta })]$, where $c\ge 0$ and the standard error is nonnegative, adding and subtracting the endpoints proves the displayed identities. Thus the endpoints must be ordered and their midpoint must equal the point estimate. Rounding should be assessed using intervals of possible unrounded values before declaring a numerical contradiction.

When this confidence interval is obtained by inverting a matching two-sided test of $\theta ={\theta }_0$, exclusion of ${\theta }_0$ agrees with rejection at the corresponding significance level, with the usual convention at boundary equality. This is a check for a specified symmetric interval procedure, such as the usual t interval for a mean difference; it does not require asymmetric intervals or intervals on transformed scales to have their midpoint at the estimate on the original scale.

Invert a family of hypothesis tests by including each parameter value in the confidence set exactly when its test does not reject. For any true value $\theta$, the events $\{\theta \notin C(X)\}$ and $\{\text{reject }{H}_{\theta }\}$ are identical. Thus size at most $\alpha$ gives coverage at least $1-\alpha$, and exact size gives exact coverage. When the accepted set is an interval, this is a confidence interval.

For the normal sample mean with unknown variance, the central Student t-distribution interval excludes zero exactly when $|\sqrt{n}\overline{X}/S|$ exceeds the same two-sided critical quantile. A two-sided p-value strictly below $\alpha$ therefore implies exclusion of zero from the matching closed interval. Boundary conventions agree up to events of probability zero in this continuous model.

For a count $s>0$ from a Poisson distribution, equal-tail inversion uses the inclusive upper-tail event $S\ge s$ for the lower endpoint and the inclusive lower-tail event $S\le s$ for the upper endpoint. Their cumulative distribution indices differ by one, as displayed. If $s=0$, the lower endpoint is zero. For $n$ observations of mean $\mu$, apply this to their total of mean $\lambda =n\mu$ and divide endpoints by $n$. The construction has at least the nominal frequentist coverage; discreteness can make it conservative.

A box contains one certainly red ball and $n\ge 1$ independently coloured extra balls, each red with probability $1/2$. Conditional on the first draw being red, a second draw without replacement is red with probability $(n+3)/(2(n+2))$. To see the effect of using a conditional probability, let $R=1+B$, where $B$ has a binomial distribution with parameters $n,1/2$. The conditional probability is $\mathbb{E}[R(R-1)]/(n\mathbb{E}[R])$: histories with more red balls are more likely to produce the first red draw and therefore receive greater weight.

Constructive solid geometry describes a solid by combining primitive regions with Boolean set union, set intersection and difference. With primitive defining functions nonpositive inside, minimum implements union and maximum implements intersection. These formulas encode membership rather than signed distance functions. Solid modellers often regularize Boolean operations to remove isolated lower-dimensional remnants.

A fillet rounds a corner or edge by inserting a tangent transition. In a planar circular construction, intersect normal lines through the prescribed tangent points to find a candidate circle centre. The two resulting radii must agree; otherwise those prescribed points are incompatible with one circular fillet. Choosing the retained side and local clipping region is part of the design: intersecting the entire original solid with the fillet disk generally removes unrelated geometry. Tangent joining need not make curvature continuous.

Joined regular parametric curves have order-$k$ geometric continuity when their local images admit compatible orientation-preserving regular parameter choices with matching derivatives through order $k$. At order one the nonzero one-sided tangent vectors must be positive scalar multiples. At order two the oriented bending must also match: equality of unsigned curvature alone does not suffice if the curves bend to opposite sides.

Joined pieces have order-$k$ parametric continuity when their positions and one-sided derivatives through order $k$ agree in the actual common parameter. Changing parameter-interval lengths changes the required derivative scale by the chain rule. For regular curves, parametric continuity implies the corresponding geometric continuity.

A boundary representation specifies a solid through its oriented bounding faces and their edge and vertex of a polyhedron incidence. Supporting planes and coordinates supply geometry; face loops and adjacency supply topology. A valid closed manifold model requires compatible orientation, two incident faces per edge and a single local fan at each vertex of a polyhedron, together with an embedded boundary.

A two-dimensional point sink of strength $Q>0$ removes volume flux $Q$ per unit depth at $z_0$. Its complex potential of a two-dimensional potential flow is the displayed expression, so $u-iv=-Q/[2\pi (z-z_0)]$. The radial velocity is $u_r=-Q/(2\pi r)$: integrating over a circle gives outward flux $2\pi r{u}_r=-Q$. This is the planar cross-section of a line sink. A passive particle satisfies $d(r^2)/dt=-Q/\pi$ and reaches the ideal point singularity in finite time. This is a flow singularity, distinct from a stable sink equilibrium of a regular dynamical system.

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.