The distance to a set is nonnegative and finite here: any fixed $y_0\in F$ gives $0\le f(x)\le |x-y_0|$. For every $y\in F$, the triangle inequality givesTaking the infimum over $y\in F$ yields $f(x)\le |x-x^{\prime }|+f(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ provesThus $f$ is a Lipschitz function with Lipschitz constant one, so it is continuous. In the epsilon-delta definition of continuity, choose $\delta =\epsilon$.

In the finite field ${\mathbb{F}}_5=\mathbb{Z}/5\mathbb{Z}$, the characteristic polynomial becomesbecause $(\lambda -3{)}^2={\lambda }^2-\lambda -1$ modulo $5$. Splitting into linear factors does not by itself guarantee diagonalizability: the repeated eigenvalue needs enough eigenvectors.

Solving $(A-3I)(x,y,z{)}^T=0$ givesso $y=3x$ and $z=4x$. Hencehas dimension one, although $3$ has algebraic multiplicity two. The geometric multiplicity is too small, and $A$ is not diagonalizable over ${\mathbb{F}}_5$. This illustrates diagonalizability depends on the field: characteristic zero gave three distinct eigenvalues, while reduction modulo $5$ merges two of them.

Over the complex numbers, the eigenvaluesare distinct. Thus distinct eigenvalues imply diagonalizability, and $A$ is diagonalizable over $\mathbb{C}$, in fact already over $\mathbb{R}$.

An explicit eigenbasis isFor either quadratic root $\lambda$, the identity ${\lambda }^2=\lambda +1$ verifies $A{v}_{\lambda }=\lambda v_{\lambda }$. Taking these three columns as $P$ gives $P^{-1}AP=\mathrm{diag}(1,{\lambda }_{+},{\lambda }_{-})$. Their distinct eigenvalues ensure linear independence.

Expansion down the last column gives the characteristic polynomialThe quadratic polynomial has roots $(1\pm \sqrt{5})/2$, which are not rational numbers. For example, the rational root theorem restricts any rational root of this monic integer polynomial to $1$ or $-1$, and neither works. If $A$ had a diagonalization over $\mathbb{Q}$, its characteristic polynomial would be a product of linear factors with rational numbers as their roots. It is not, so $A$ is not diagonalizable over $\mathbb{Q}$.

The span $U$ is the coordinate plane $\{(x_1,x_2,0,0{)}^T\}$. Consequently its annihilator of a vector subspace is $U^{\circ }=\mathrm{span}\{{\epsilon }_3,{\epsilon }_4\}$. The dual map is composition with $T$, soIt follows thatEquivalently, $T^{\prime }(a{\epsilon }_3+b{\epsilon }_4)$ sends $x$ to $a{x}_3$.

Let ${\epsilon }_i$ denote the coordinate linear functional dual to the $i$th vector of the standard basis. Writing a vector in the supplied basis givesSuccessive differences recover the coefficients. Hence the dual basis isThese formulas directly give $v_i^{\prime }(v_j)={\delta }_{ij}$ and exemplify the dual basis of cumulative coordinate sums.

For the basis $e_1,\dots ,e_n$, define the dual basis byUniqueness of the coordinates makes each $e_i^{\prime }$ well-defined and linear. If ${\sum }_i{a}_i{e}_i^{\prime }=0$, evaluation at $e_j$ gives $a_j=0$, so these linear functionals are linearly independent. For any $\lambda \in V^{\prime }$, evaluation on each basis vector givesBoth sides agree on a basis, hence on every linear combination. Thus they also span the dual space and are a basis of it.

The annihilator of a vector subspace is the vector subspace of the dual space consisting of the linear functionals that vanish on $U$:It is a vector subspace because every linear combination of functionals vanishing on $U$ still vanishes there.

Along the circle, $z=z_0+r{e}^{it}$ and $dz=ir{e}^{it}dt$. Expanding the nonholomorphic integrand givesAfter multiplication by $ir{e}^{it}$, all nonconstant integer-frequency exponentials have zero integral over $[0,2\pi ]$. The remaining constant term is $i{r}^2{z}_0$. ThereforeThis direct parametrization avoids applying a holomorphic-antiderivative argument to $|z{|}^2$, which is not holomorphic.

An entire antiderivative is $F(z)=z\sin z+\cos z$, since differentiation cancels the sine terms and leaves $F^{\prime }(z)=z\cos z$. The curve runs from $-i$ to $i$. Both terms of $F$ are even, so $F(i)=F(-i)$, and the complex line integral gives

The rational cut is both open and closed in $\mathbb{Q}$. It is relatively open as $\mathbb{Q}\cap (-\infty ,\sqrt{2})$. Since $\sqrt{2}$ is irrational, its relative complement is $\mathbb{Q}\cap (\sqrt{2},\infty )$, also relatively open. Thus this cut is a clopen set in the subspace topology of the rationals.

In the subspace topology, $B$ is relatively open in $A$ if $B=A\cap U$ for an open subset $U$ of the ambient Euclidean space. Equivalently, every $b\in B$ has a radius $\epsilon >0$ with $A\cap B_{\epsilon }(b)\subseteq B$. It is relatively closed if its complement $A\setminus B$ is relatively open, equivalently $B=A\cap F$ for an ambient closed set $F$.

For the upper half of the open disk, $B$ is closed in $A$, but not open in $A$. It is the intersection with the closed half-plane $\mathrm{Im}z\ge 0$. A relative neighborhood of zero contains points of negative imaginary part, so relative openness fails.

For open sets in $\mathbb{C}$, the answers are $X_1$: no; $X_2$: no; $X_3$: yes. Every complex disk centered at a real point contains nonreal points, so a nonempty subset of the real axis cannot be open in the complex plane. The empty set remains open.

For open sets in $\mathbb{R}$, the answers are $X_1$: yes; $X_2$: no; $X_3$: yes. Every point of the positive ray has a small interval contained in it, while zero is not an interior point of the nonnegative ray. The empty set is open by the topology axioms.

For closed sets in $\mathbb{R}$, the answers are $X_1$: no; $X_2$: yes; $X_3$: yes. The first interval omits its limit point zero. The second contains its finite limit points, and the empty set is closed in every topology.

Parametrize the horizontal line as $z=t+ib$, $t\in \mathbb{R}$. Then $e^z=e^t{e}^{ib}$, and $e^t$ takes every positive real value exactly once. The image isThis is the ray with argument $b$ modulo $2\pi$, excluding its endpoint zero. As $t$ increases, the image moves outward from arbitrarily small to arbitrarily large radius.

Parametrize the vertical line as $z=a+it$, $t\in \mathbb{R}$. Then $e^z=e^a(\cos t+i\sin t)$, whose modulus is $e^a$ and whose argument takes every value. The image isIncreasing $t$ traverses the circle counterclockwise, repeating with period $2\pi$. There is no injectivity on the full line.

This is the wave equation, again taking $y$ as time. Zero is not unique: for $k=\pi /2$,satisfies $u_{xx}=u_{yy}$ and all the stated boundary conditions. The initial displacement is zero but the initial velocity is not prescribed.

A sufficient extra condition is $u_y(x,0)=0$. For the resulting homogeneous data use wave energyThe equation and integration by parts give $E^{\prime }(y)=[u_x{u}_y{]}_0^1=0$. At $x=1$ the factor $u_x$ vanishes, and at $x=0$ the factor $u_y$ vanishes because $u(0,y)=0$. The initial displacement implies $u_x(x,0)=0$, so the extra initial velocity condition gives $E(0)=0$. Consequently $u_x=u_y=0$ and $u=0$. With the stated boundary regularity, integrate the local energy conservation law over inset rectangles and let their edges approach the boundary; the energy and flux terms converge by continuity of the first derivatives.

This is Laplace's equation. Zero is not unique: with $k=\pi /2$,is nonzero and harmonic. It vanishes on $y=0$ and $x=0$, while $u_x(1,y)=k\cos (k)\sinh (ky)=0$.

A sufficient additional condition is $u(x,1)=0$ on the top edge. Green's first identity then givesOn the bottom, left and top edges $u=0$; on the right edge the normal derivative is zero. Thus the gradient vanishes and $u$ is constant; the bottom value makes that constant zero. For $C^2(D)\cap C^1(\overline{D})$, apply the identity on inset rectangles and pass to the boundary using continuity of the function and its gradient.

This is the heat equation, with $y$ as time. The stated zero solution is unique. Define $E(y)=\frac{1}{2}{\int }_0^{1}u(x,y{)}^{2}dx$. For $0<y<1$, integration by parts and $u_y=u_{xx}$ giveThe boundary term is zero because $u(0,y)=0$ and $u_x(1,y)=0$. Also $E(0)=0$. Nonnegativity forces $E(y)=0$, and continuity gives $u=0$ throughout the square. The unknown final-time boundary values are determined by evolution and need not be prescribed. Under the stated regularity the integrations can first be made on an interior interval and then passed to the endpoints using continuity of $u,u_x,u_y$.