Let $\{O_i\}$ be an open cover of $Z$ in its subspace topology. Write $O_i=Z\cap U_i$ with each $U_i$ open in $X$. Since $Z$ is a closed set, $X\setminus Z$ is open, and $\{U_i\}\cup \{X\setminus Z\}$ is an open cover of the compact space $X$. A finite subcover restricts to a finite subcover of $Z$. Hence the closed subset $Z$ is compact.

If $\{V_j\}$ is an open cover of $f(Z)$ in its subspace topology, the inverse images under the restricted continuous map $f{|}_Z$ form an open cover of $Z$. Take a finite subcover; the corresponding $V_j$ cover $f(Z)$. Thus the continuous image $f(Z)$ is compact. Neither argument assumes a Hausdorff space or asserts that a compact set must be closed set in an arbitrary topological space.

Induct on $n$ to prove the stronger assertion that any submodule $M\subseteq R^n$ is a free module of rank of a free module at most $n$. For $n=0$ this is immediate. Project onto the first coordinate by $\pi :R^n\to R$. Its image $\pi (M)$ is an ideal of the principal ideal domain, so $\pi (M)=(a)$.

If $a=0$, then $M$ embeds in the remaining $n-1$ coordinates and the induction hypothesis applies. Otherwise choose $v\in M$ with $\pi (v)=a$. Put $N=M\cap \ker \pi$, a submodule of $R^{n-1}$, which is free by induction. For every $m\in M$, there is $b\in R$ with $\pi (m)=ba$, so $m-bv\in N$. If $bv\in N$, then $ba=0$; the integral domain property and $a\ne 0$ give $b=0$. ThereforeAppending $v$ to a basis of $N$ proves that $M$ is a free module of rank of a free module at most $n$. This proof of the submodule theorem for free modules over a principal ideal domain does not assume in advance that $M$ is a finitely generated module.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

For the usual undirected graph with at least two vertices, a simple random walk chooses a neighbor uniformly at each step, so $P_{ij}=1/d_i$ when $i$ and $j$ are adjacent and zero otherwise, where $d_i$ is the degree of a vertex. The handshaking lemma gives ${\sum }_i{d}_i=2|E|$. Thus the stationary distribution of a graph random walk isFor adjacent $i,j$, ${\pi }_i{P}_{ij}=1/(2|E|)={\pi }_j{P}_{ji}$, and for nonadjacent pairs both sides vanish. By detailed balance, this is an invariant distribution and the walk started from it is a reversible Markov chain. Connectedness makes the finite walk an irreducible Markov chain, giving uniqueness of its invariant distribution. No aperiodicity assumption is needed for uniqueness, so a bipartite graph is allowed.

If the connected graph has a single isolated vertex, the usual neighbor-choice formula is undefined. With the natural absorbing convention $P_{11}=1$, its unique invariant distribution is ${\pi }_1=1$ and it is reversible. The degree-normalization formula presupposes at least one edge.

A reversible Markov chain has a stationary law $\pi$ for which every finite path has the same distribution as its time-reversed path:For a transition matrix $P$, the equivalent detailed balance condition is ${\pi }_i{P}_{ij}={\pi }_j{P}_{ji}$ for all $i,j$. It implies $\pi P=\pi$ by summing over $i$. Multiplying the balance identities along a path shows equality of its forward and reverse probabilities when $X_0\sim \pi$; conversely, taking paths of length one gives those identities.

The Born rule gives the probability of $E_n$ as the squared coefficient of the initial state in the energy eigenstate basis. Direct integration givesThe strict energy cutoff corresponds to $n^2<5$, so only $n=1,2$ contribute. Both coefficients equal $2/\pi$, and thereforeThe discontinuous initial wavefunction is a legitimate normalized $L^2$ state for this spectral probability calculation, although it is not in the Hamiltonian's differentiability domain; finite mean energy is not required to compute this probability.

Inside the infinite square well, the time-independent Schrödinger equation is $-({\hslash }^2/(2m)){\psi }^{\prime \prime }=E\psi$. The infinite exterior potential imposes $\psi (0)=\psi (a)=0$. For $E>0$, the first condition leaves a sine and the second requires $ka=n\pi$. A zero or negative energy gives only the zero solution under these conditions. The normalized energy eigenstates and eigenvalues areThe wavefunctions vanish outside the well, and a constant phase is immaterial. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$.

Let $c>0$ denote the stated unit cost of cylindrical wall, and assume $V>0$. The circular cylinder and hemisphere give the volume constraint and total costOnly the walls and cap have specified costs, so no separate base cost is included. Eliminating $h$ giveswith $0<r\le (3V/(2\pi ){)}^{1/3}$ from $h\ge 0$. Now $C^{\prime }(r)=-2cV/r^2+(16/3)\pi cr$ vanishes at $r^3=3V/(8\pi )$. Moreover $C^{\prime \prime }(r)=4cV/r^3+(16/3)\pi c>0$ and the stationary point is feasible, so it is the unique global minimum. The optimal dimensions are

For the boundary constraint $g(x,y)=x^2+y^2-1=0$, the Lagrange multiplier equations for $z=xy$ areNeither coordinate can vanish at a stationary point. Substitution gives $4{\lambda }^2=1$. For $\lambda =1/2$, $x=y=\pm 1/\sqrt{2}$ and $z=1/2$; for $\lambda =-1/2$, $x=-y=\pm 1/\sqrt{2}$ and $z=-1/2$. The constraint circle is compact, its gradient is nowhere zero, and every extremum must occur among these stationary points. ThusThe paired signs in each expression are linked. The inequality $2|xy|\le x^2+y^2=1$ also confirms the global values.

If $X$ is Hausdorff, every pair $x\ne y$ has disjoint open neighborhoods $U$ and $V$. The rectangle $U\times V$ is an open neighborhood of $(x,y)$ disjoint from the diagonal. Hence every point of its complement is interior, so the diagonal is a closed set.

Conversely, if the diagonal is closed, its complement is open in the product topology. For $x\ne y$, the rectangle-basis property gives open $U\ni x$ and $V\ni y$ with $U\times V$ disjoint from the diagonal. If $z\in U\cap V$, then $(z,z)$ lies in that rectangle, a contradiction. Thus $U\cap V=\varnothing$, giving the Hausdorff separation condition. ConsequentlyThe argument includes spaces with zero or one point, for which both conditions hold.

The product topology has as a basis of a topology all sets $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Open sets are arbitrary unions of these rectangles. The projection maps are $p(x,y)=x$ and $q(x,y)=y$. For any open $U\subset X$ and $V\subset Y$,These are open rectangles, including the empty case. Taking inverse images of open sets therefore proves that both projection maps are continuous functions.

Let $x_n=(\pi /2+2\pi n{)}^{-1}$. The points $p_n=(x_n,1)$ lie on the specified graph and converge in ${\mathbb{R}}^2$ to $(0,1)$. They form a Cauchy sequence in the induced metric, but their only possible limit is absent from the space. Adding $(0,0)$ does not supply that missing limit. Hence the specified space is not complete.

For the unheaded request on $\mathbb{R}$, use the arctangent pullback metricPositivity, symmetry and the triangle inequality follow from the ordinary distance, and injectivity of $\arctan$ gives $d(x,y)=0$ only if $x=y$. The sequence $n$ is Cauchy for this metric, because $\arctan n\to \pi /2$, but cannot converge to any finite real $x$: that would require $\arctan x=\pi /2$. Thus this metric makes $\mathbb{R}$ incomplete, even though it induces its usual topology.

The rational sequence $q_n=10^{-n}\lfloor 10^n\sqrt{2}\rfloor$ satisfies $0\le \sqrt{2}-q_n<10^{-n}$. It is therefore a Cauchy sequence in the induced metric on $\mathbb{Q}$. Any limit in $\mathbb{Q}$ would also be its real limit, which is $\sqrt{2}\notin \mathbb{Q}$. ThusCompleteness concerns limits lying in the space, so being a dense subset of $\mathbb{R}$ does not suffice.

A metric space is a complete metric space when every Cauchy sequence converges to a point of that space: for each $\epsilon >0$, the condition $d(x_m,x_n)<\epsilon$ for all sufficiently large $m,n$ must imply existence of $x$ with $d(x_n,x)\to 0$.

A Cauchy sequence in $[0,1]$ is also Cauchy in $\mathbb{R}$, so it converges there by the assumed completeness. Its limit remains between zero and one, since inequalities are preserved under limits. ThereforeThis is an instance of the fact that a closed subset of a complete metric space is complete with its induced metric.

In this consecutive-number antisymmetric game, with the row player's number denoted $i$, the payoff matrix isIt is an antisymmetric matrix, so the matrix game has value zero for every $n\ge 3$. For $n=3$,satisfies $Ap=0$ and $p^{T}A=0$, proving optimality for both players.

For any larger $n$, extend the same strategy by zero probabilities:Indeed, the first three coordinates of $Ap$ are zero. For $i=4$, $(Ap{)}_4=-2/4-2/2+1/4=-5/4$; for $i\ge 5$, all three supported numbers are at least two below $i$, so $(Ap{)}_i=-2$. Hence $Ap\le 0$ and, by antisymmetry, $p^{T}A\ge 0$. The column player using $p$ prevents positive payoff, while the row player using $p$ prevents negative payoff, a mixed-strategy optimality certificate for a matrix game. This explicitly checks all the additional strategies rather than assuming the three-number subgame remains optimal.

For a real antisymmetric matrix $A$, $p^{T}Ap=0$ for every real vector $p$. Thus for each row strategy $p$, the column player could choose $q=p$, giving ${\min }_q{p}^{T}Aq\le 0$. Hence the row security value is at most zero. Conversely, for each column strategy $q$, the row choice $p=q$ gives ${\max }_p{p}^{T}Aq\ge 0$, so the column security value is at least zero. The minimax theorem equates these values, provingThe choices matching the opponent in this proof establish inequalities for security levels; they do not by themselves specify a single universal optimal strategy.

In a matrix game, entry $a_{ij}$ is the row player's payoff, and the column player's payoff is $-a_{ij}$. A mixed strategy is a probability vector over the available rows or columns. With strategies $p,q$, the row payoff is $p^{T}Aq$. Its optimal mixed strategy maximizes its guaranteed payoff, while the column player minimizes the largest row payoff:The equality is the finite minimax theorem. Equivalently the row player maximizes ${\min }_j(p^{T}A{)}_j$, and the column player minimizes ${\max }_i(Aq{)}_i$. Each optimal strategy guarantees its corresponding bound against every opponent strategy; for the column player's own negative payoff this is also a maximin strategy.

Write the one-column design as $x\in {\mathbb{R}}^n\setminus \{0\}$ and $S=x^{T}x$. Maximizing the normal linear model likelihood first gives $\hat{\beta }=x^{T}Y/S$. Its minimized residual sum of squares is $R=\Vert Y-x\hat{\beta }{\Vert }^2$. For variance $v>0$, the remaining log likelihood is $-\frac{n}{2}\log v-R/(2v)$ plus a constant, whose maximum is atHere $R>0$ almost surely because $n>1$ and the true variance is positive. In the exceptional data case $R=0$, the likelihood is unbounded as $v\downarrow 0$, so there is no positive-variance maximum.

Let $P=x{x}^T/S$, the orthogonal projection matrix onto the design column. Then $\hat{\beta }-\beta =x^T\epsilon /S$ and $Y-x\hat{\beta }=(I-P)\epsilon$. These are orthogonal linear projections of an isotropic multivariate normal distribution, hence independent. An orthonormal coordinate system adapted to $x$ makes the residual squared length a sum of $n-1$ independent squared standard normals. Thus the joint distribution of least-squares and variance estimators is the product lawIn particular the maximum likelihood variance estimator is biased: $\mathbb{E}{\hat{\sigma }}^2=(n-1){\sigma }^2/n$; the unbiased residual estimator uses denominator $n-1$.

The Gauss-Markov theorem states that with full-column-rank design, mean-zero errors and covariance ${\sigma }^{2}I$, the ordinary least squares estimator is the best linear unbiased estimator: its covariance is no larger than any other such estimator's in the Loewner order. Normality is unnecessary for that theorem.

In the given normal linear model, maximizing the likelihood function is equivalent to minimizing $\Vert Y-X\beta {\Vert }^2$. The normal equations and full rank giveFor a direct variance comparison, write ${\beta }^{\ast }=AY$. Unbiasedness for every $\beta$ requires $AX=I_p$. Set $B=(X^{T}X{)}^{-1}{X}^T$ and $D=A-B$, so $DX=0$. The cross covariance terms vanish because $B{D}^T=(X^{T}X{)}^{-1}(DX{)}^T=0$. ThereforeThis proves the requested comparison for every contrast $t$ without relying only on the theorem's name.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.

The charge feels the induced field, equivalently the field of the image charge, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so Coulomb's law givesIts magnitude is $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.