The norm induced by a real inner product isPositive definiteness and homogeneity follow from the inner product axioms; Cauchy-Schwarz gives the triangle inequality by expanding $\Vert u+v{\Vert }^2$.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

Independence of the two samples makes their sample means independent normal random variables. A difference of independent normal random variables is normal, with the means subtracted and the variances added. Thus

A state $i$ is recurrent if the probability of returning to $i$ at a strictly positive time, starting at $i$, equals one. It is transient if that probability is less than one. If $i$ and $j$ communicate, choose $r,s$ with $p_{ji}(r)>0$ and $p_{ij}(s)>0$. Chapman-Kolmogorov givesIf $i$ is recurrent, summing over $n$ makes the right-hand side diverge, so the allowed recurrence criterion proves $j$ recurrent. Interchanging $i,j$ proves recurrence is shared throughout a communicating class, and therefore transience is shared too.

For the simple random walk, an odd number of steps cannot return to its starting point, andFor $p=1/2$ the return series diverges like $\sum n^{-1/2}$, proving recurrence. For $0<p<1$, $p\ne 1/2$, its tail is bounded by a convergent geometric series since $4p(1-p)<1$, proving transience. At $p=0$ or $1$ the walk is deterministic and never returns. Translation invariance transfers the conclusion to every state. Thus recurrence holds exactly when $p=1/2$.

Condition on $X_m=k$ and apply the Markov property and time homogeneity:Thus the transition matrices satisfy $P(m+n)=P(m)P(n)$; since $P(1)=P$ and $P(0)=I$, induction gives $P(n)=P^n$. Averaging over the initial state givesIf $\mu P=\mu$, induction gives $\mu P^n=\mu$, so this probability equals ${\mu }_i$ for every $n$. This is stationarity of the initial distribution.

Reflection in the line $y=-x$ is $Rz=\overline{iz}=-i\bar{z}$. The printed expression has two conjugation bars: $w(z)+\overline{w(\overline{iz})}$. On the line $Rz=z$, this is $w(z)+\overline{w(z)}$, which is real; hence its stream function vanishes there. Applying this reflection to the dipole givesConsequentlyThe image lies at $-i\bar{a}$, with moment $-i\bar{m}$, on the opposite side of the wall.

Reflection in the real axis replaces a dipole at $a$ with an image at $\bar{a}$ of moment $\bar{m}$. Reflection in the imaginary axis places the image at $-\bar{a}$ with moment $-\bar{m}$. Applying both reflections gives the image at $-a$ with moment $-m$. ThusFor example, on the real axis the first two terms are conjugates and the last two are conjugates. On the imaginary axis the first and third terms are conjugates, as are the second and fourth. The potential is real on each wall, so both walls are streamlines. The only singularity in the open first quadrant is the original dipole.

The Milne-Thomson circle theorem says that, when $f$ is a complex potential analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle isThe reflected expression is analytic in the exterior: conjugating a local power series turns its argument into $R^2/z$. It creates only interior image singularities. On $|z|=R$ we have $R^2/\bar{z}=z$, so $W=f+\bar{f}$ is real. The stream function is therefore constant on the circle, which is the impermeability condition. At infinity the reflected term is $\overline{f(0)}+O(z^{-1})$, so it changes the velocity by $O(z^{-2})$ and leaves the imposed far flow unchanged. These observations prove the theorem, up to the irrelevant additive potential constant and with no extra circulation introduced.

With $a=z_1$, $|a|>1$, and unit radius, it givesThe image dipole is at $1/\bar{a}$ with moment $-\bar{m}/{\bar{a}}^2$; writing the image term this way changes only an additive constant.

The Lagrange multiplier condition says that at a constrained local extremum of a continuously differentiable objective on a regular surface $h=0$, its gradient is normal to the tangent plane: $\nabla f=\lambda \nabla h$ for some scalar multiplier.

First the box must be centred at zero. If the two face coordinates in one direction are $u_i,v_i$, subtracting the ellipsoid equations for two vertices differing only in that coordinate gives $u_i^2=v_i^2$. A nondegenerate box has $u_i\ne v_i$, hence $u_i=-v_i$. Its positive half-sides $x,y,z$ therefore satisfy $x^2/a^2+y^2/b^2+z^2/c^2=1$ and its volume is $8xyz$.

The feasible closed octant is compact, the boundary gives zero volume, and some interior box has positive volume, so the maximum is interior. Apply multipliers to $\log (xyz)$ there:Thus the three squared scaled coordinates are equal and sum to one. The unique positive solution gives

The defining function $h(x)=x_1^2/a^2+x_2^2/b^2+x_3^2/c^2-1$ is smooth and has gradient $2(x_1/a^2,x_2/b^2,x_3/c^2)$. This cannot vanish on its zero level, since the origin does not lie there. The regular level-set theorem therefore makes $M$ a two-dimensional submanifold, with tangent space $\ker Dh(x)$ and normal space

A two-dimensional embedded submanifold is locally flattened by a continuously differentiable coordinate diffeomorphism of an ambient open neighbourhood onto a coordinate plane, or equivalently locally admits a regular two-parameter chart. A tangent vector at $x$ is the velocity ${\gamma }^{\prime }(0)$ of a differentiable curve in $M$ through $x$. All such vectors form the two-dimensional tangent space $T_{x}M$, the image of the differential of a regular parametrization. Normal vectors form its orthogonal complement $(T_{x}M{)}^{\perp }$ in ${\mathbb{R}}^3$, which has dimension one.

The claim is false for an arbitrary codomain $Y$. Let $X=[0,1]$ with its usual topology, let $Y$ be the Sierpiński space with open sets $\varnothing ,\{1\},Y$, and define $f(0)=0$, $f(x)=1$ for $x>0$. This is a continuous map: the only nontrivial open-set preimage is $f^{-1}(\{1\})=(0,1]$, which is relatively open in $X$.

The sets $K_n=[0,1/n]$ are nonempty, nested and compact. ButThus the extra generality in the printed codomain changes the assertion materially. This is the failure of nested compact image intersection without closed fibers.

If the codomain is a T1 space, the continuous images of nested compact intersections theorem does give the desired equality. The forward inclusion holds for every map. For the reverse, take $y\in {\bigcap }_{n}f(K_n)$ and set $F_n=K_n\cap f^{-1}(\{y\})$. The fibers are closed by continuity and the T1 property, so the $F_n$ are nested nonempty closed subsets of the compact set $K_1$. The finite intersection property gives $x\in {\bigcap }_n{F}_n$, with $f(x)=y$. It is enough that these particular fibers be closed in $K_1$; a Hausdorff codomain, as in the native self-map formulation, is a familiar sufficient assumption.

In the usual real line, take $\boxed{F_n=[n,\infty )}$ for $n\ge 1$. The sets are closed, nonempty and nested, but no real number belongs to all of them.

Take radius $r>0$. Part (a) gives one containment. For a point $x$ with $\Vert x-a\Vert \le r$, define $x_t=a+t(x-a)$ for $0\le t<1$. Then $\Vert x_t-a\Vert =t\Vert x-a\Vert <r$, while $x_t\to x$ as $t\uparrow 1$. Hence every point of the closed ball belongs to the closure of the open ball, proving equality. Positive radius is the standard open-ball convention; a zero-radius open ball would be empty.

Intersect the ambient balls with $E$. All points on the vertical arm have distance at least $1/2$ from the centre, so the open ball isThe closed ball also contains the endpoints of that segment and the lower half of the vertical arm:The closure of the open ball in $E$ is just $[0,1]\times \{0\}$. Both endpoints are limits of points in the open segment, whereas a vertical point of positive height has a small enough neighbourhood disjoint from it. Thus every point $(0,y)$ with $0<y\le 1/2$ witnesses strict containment.

The condition $\max (|x-1/2|,|y|)\le 1/2$ gives the closed squareIts four edges and corners are included, as shown in the left panel of the figure below.