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Factor the denominator as $(z^2+1)(z^2+4)$ and integrateover $[-R,R]$ followed by the upper semicircle, for $R>2$. The enclosed simple poles are $i$ and $2i$. The local quotient formula from part (a) givesOn this arc, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+1|\ge R^2-1$, $|z^2+4|\ge R^2-4$. Thus the estimation lemma givesOn the real line, $|F(x)|\le 1/[(x^2+1)(x^2+4)]$, so the improper integral is absolutely convergent. Hence the residue theorem yieldsTaking real parts is justified by absolute convergence. ThereforeThe imaginary part vanishes as a further check, since the sine integrand is odd.

The circle is path-connected, using circular arcs, and hence connected. Its continuous image is connected by part (ii). A connected subset of the real line is an interval: if it contained values on opposite sides of a missing value $c$, intersection with $(-\infty ,c)$ and $(c,\infty )$ would separate it. Thus $f(C)$ is an interval. It is also compact, so it is a closed bounded interval, possibly reduced to one point.

The family contains the empty set and the entire real line. The intersection of two proper nonempty members isso it is closed under finite intersection. For an arbitrary nonempty family of proper rays, put $c={\inf }_i{a}_i$. If the thresholds are unbounded below, the union is $\mathbb{R}$; otherwise the union is $(c,\infty )$. In the latter case $c$ itself is in no ray, and every point greater than $c$ lies above some threshold, proving the equality. Empty unions and unions involving $\mathbb{R}$ cause no difficulty. Thus arbitrary unions also belong to the family, establishing the topology axioms.

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.

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.

No. Take $f_n=1_{[n,n+1]}$ on the real line. At every fixed point it eventually vanishes, so $f=0$, while $\int f_n=1$ for every $n$. Thus the limit of the integrals is one and the integral of the pointwise limit is zero. Nonnegativity alone cannot replace monotonicity or an integrable domination hypothesis.

In the usual real line take $\boxed{F_n=[n,\infty )}$. Each set is nonempty and closed, and $F_{n+1}\subseteq F_n$. No real number lies in them all, because the positive integers are unbounded. Thus their intersection is empty. These sets are not compact.

The connected subsets of the real line are exactly the intervals, including singletons, unbounded intervals and the empty set. Equivalently, whenever $x<y$ belong to the set, every $z$ with $x<z<y$ also belongs to it. Endpoints may or may not be included.

The stationary Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)+V\psi =E\psi$. A nonzero bound-state function is square-integrable on the real line and can be scaled to ${\int }_{\mathbb{R}}|\psi {|}^{2}dx=1$. Odd parity means $\psi (-x)=-\psi (x)$.

Set $k=\sqrt{2m(V_0+E)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$, both positive. Oddness and decay giveContinuity at $a$ gives $B=A\sin (ka)$, and derivative continuity gives $-\kappa B=Ak\cos (ka)$. A nonzero solution cannot have $\sin (ka)=0$, since that would force both the value and derivative there to vanish. Dividing therefore yields $\kappa =-k\cot (ka)$. With $\eta =\kappa a$ and $\xi =ka$,The matching at $-a$ follows automatically by odd parity.

In the required sketch, the second equation is a first-quadrant circle arc of radius $R=\sqrt{2m{V}_0{a}^2}/\hslash$. The curve $-\xi \cot \xi$ starts with limiting value $-1$ at zero, rises to zero at $\pi /2$, and then rises to $+\infty$ as $\xi \uparrow \pi$. Thus a positive-$\eta$ intersection in this range must have $\xi >\pi /2$. The accompanying plot shows the circle below threshold and one above it.

Consequently $R<\pi /2$ rules out all odd states: the first positive branch already requires that radius, and later branches require still larger $\xi$. This is precisely $\boxed{2m{V}_0<(\pi \hslash /(2a){)}^2}$. At equality the intersection has $\eta =0$ and would not decay at infinity, so it too is not a normalizable bound state. An odd state first appears only for a strict increase beyond threshold.

Put $A={\sum }_i{x}_i$, $B={\sum }_i{x}_i^2$ and $M_2=B/n$. The normal distribution model givesand, up to a constant independent of $\mu$, its log-likelihood function isThe score function and its zero equation areFor $B>0$, this quadratic equation has exactly one positive root, giving the maximum-likelihood estimatorThe numerator $B-A\mu -n{\mu }^2$ of the score function is positive between its negative and positive roots and negative above its positive root. Thus the likelihood function increases up to $\hat{\mu }$ and decreases afterwards: this is its unique global maximum on $(0,\infty )$. Also, $\ell (\mu )\to -\infty$ at both ends. The all-zero observation vector has probability zero; on that exceptional vector $L(\mu )\propto {\mu }^{-n}$ and there is no attained maximum.

The second derivative and the relevant moments areTherefore the expected Fisher information isThe support of a probability distribution is the whole real line, independent of the positive statistical parameter. The law of large numbers gives $\overline{X}\to \mu$ and $M_2\to 2{\mu }^2$ in probability, so the displayed maximum-likelihood estimator is a consistent estimator. The regular asymptotic normality of a maximum likelihood estimator now yieldsThus the requested leading asymptotic variance on the unscaled estimator's scale is ${\mu }^2/(3n)$. The normal mean proportional to standard deviation model provides a reusable version of this calculation.

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

The two conditions for a basis of a topology both hold. First, every real number $x$ belongs to $[x,x+1)$, so the proposed basic intervals cover the real line. Second, ifthen $\max (a,c)\le x<\min (b,d)$, andis itself one of the proposed basic intervals. Thus every point in an intersection of two basic sets has a basic neighbourhood contained in that intersection.

Consequently the arbitrary unions of members of $\mathcal{B}$, including the empty union, form a topology $\mathcal{T}$. The basis of a topology conditions give closure under finite intersections; arbitrary unions are built into the definition. This is the lower limit topology, also called the Sorgenfrey line.

Let $U$ be open in the Euclidean topology on the real line, and let $x\in U$. There is an $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq U$. The basic open set $[x,x+\epsilon )$ contains $x$ and is contained in $U$. Taking the union of these basic neighbourhoods over $x\in U$ shows that $U\in \mathcal{T}$. ThusIn fact this is a strictly finer topology: $[0,1)$ is open in $\mathcal{T}$ but not in the Euclidean topology, since no usual open ball about $0$ is contained in it.

The usual real line is connected. In contrast,is a separation of a topological space. Both pieces are nonempty and disjoint; the first is open by part (c), and the second is the open union ${\bigcup }_{n\ge 1}[0,n)$. Thus the Sorgenfrey line is not connected. Since a homeomorphism preserves connectedness, the two topological spaces are not homeomorphic. Merely showing that their topologies differ would not by itself exclude some other homeomorphism between them.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

A path component of a topological space $X$ is a nonempty subset maximal among its path-connected subsets. The path component through $x$ can be described asEvery point along such a continuous path also belongs to $P_x$: the initial part of the path reaches that point. Thus the paths witnessing membership stay inside $P_x$. This makes path components useful for deciding which allowed positions can be reached from one another by continuous motion.

Every path component lies inside one connected component, because every path-connected space is connected. Thus each connected component is a union of path components. In a locally path-connected space the two classifications agree, by connected locally path-connected spaces are path-connected. For a subset of the real line with its usual subspace topology, the classifications also agree: every connected component is a real interval, and interpolation supplies a continuous path within it.

The topologist's sine curve shows why the distinction matters in general. Use specifically the closed versionwith the subspace topology from the usual Euclidean plane. It has one connected component, $T$, but two path components, $G$ and $L$. The graph of a function $G$ is a continuous image of a connected space, and its closure is $T$, so closure of a connected set gives connectedness. Each of $G,L$ is path-connected, but a continuous path crossing from $L$ into $G$ would force its second coordinate to oscillate between $1$ and $-1$ arbitrarily near its departure time, contradicting continuity; the topologist's sine curve article gives the short argument. Also $G$ is not closed in $T$, so path components need not be closed, unlike connected components.^[ref]

Being joined by a continuous path is an equivalence relation. Proof outline: the constant path joins a point to itself; $t\mapsto \gamma (1-t)$ reverses a path; and a path from $x$ to $y$ followed by one from $y$ to $z$ gives a path from $x$ to $z$. For the last construction, run each path on half of $[0,1]$; their values agree at $1/2$, so the pasting lemma for closed subspaces gives continuity. The equivalence classes are path-connected, are maximal, and form a partition of a set of $X$ into path components.^[ref]

For $T>0$, values of a function $f$ on a half-open interval $[a,a+T)$ determine its periodic extension to the real line: choose the unique integer $k$ with $x-kT\in [a,a+T)$ and set $F(x)=f(x-kT)$. The resulting periodic function has period $T$. If the endpoint limits of $f$ differ, the extension has a jump discontinuity at every period boundary. Under the usual piecewise-smooth convergence hypotheses, its Fourier series takes the average of the two one-sided limits there; an assigned value at an isolated endpoint does not change the Fourier coefficients.

Define the indicator function of the rational numbers on the real line by $f(x)=1$ for rational $x$ and $f(x)=0$ otherwise. Every rational shift preserves rationality and irrationality, so every nonzero rational number is a period of a function. Conversely, an irrational shift sends $x=0$ from value one to value zero, so it is not a period. The period group of a function is exactly $\mathbb{Q}$, and the positive periods $1/n$ approach zero without a smallest one. The function is nonconstant and discontinuous everywhere; this is why continuity matters in continuous periodic functions have a least positive period.

A power law is a relation $f(x)=C{x}^p$, with fixed constants $C$ and $p$. For positive $x$, any real number $p$ is allowed. Its useful feature is a simple scaling rule:The same proportional change of input produces the same proportional change of output, independently of the starting value. The exponent controls that response; the coefficient fixes the magnitude.

For example, the area of a circle is $A(r)=\pi r^2$. Doubling the radius multiplies its area by four, and tripling it multiplies the area by nine. The volume of a sphere is $V(r)=4\pi r^3/3$: the same changes multiply its volume by eight and twenty-seven. The distinction explains why similarly shaped objects do not preserve their ratio of surface area to volume when their size changes.

When the output is positive, a logarithm turns the power law into a straight-line relation. Using positive reference values $x_0$ and $f_0=f(x_0)$,These ratios are dimensionless quantities, so the formula also applies to physical measurements with units. Plotting both ratios on a logarithmic scale gives a line of slope $p$. Changing a unit changes the numerical coefficient, while the exponent stays the same. Positive $p$ describes increasing values; negative $p$ describes decreasing values; $p=0$ gives a constant. The zero coefficient is a degenerate case with no useful logarithmic slope.

Exact scaling is stronger than regular variation. The function $x^2\log x$ has a doubling ratio tending to four, but its logarithmic correction keeps growing. The diagram contrasts it with the exact power law $x^2$.

A power law for a probability density and a power law for a tail probability have different exponents. For a Pareto distribution with $\alpha >0$ and lower endpoint one,The tail probability has exponent $-\alpha$; differentiating it and changing sign gives the probability density exponent $-\alpha -1$. Confusing them changes the interpretation of a fitted exponent. The normalizing coefficient and lower endpoint also matter: a decreasing power law is not automatically a valid probability density over the entire positive real line.

The appearance of a line over a finite logarithmic range is evidence to investigate, rather than a proof of an exact power law or its mechanism. In their study of power-law distributions, Clauset, Shalizi and Newman combine maximum likelihood estimation, goodness-of-fit tests and comparisons with alternative models; their abstract explains why ordinary least-squares fitting can give misleading results for tail data. See Aaron Clauset, Cosma Rohilla Shalizi and M. E. J. Newman, Power-law distributions in empirical data, SIAM Review 51 (2009), 661–703. The scaling rule is mathematically exact when it defines a model; whether a dataset supports that model is a question of statistical inference.

For a non-dividend-paying traded asset $dS=\mu Sdt+\sigma dW$ and bank account $dB=rBdt$, delta hedging sets the share holding to $V_s$. Its bond value is $V-s{V}_s$. Itô's formula and the self-financing condition give the displayed pricing equation on the whole real line. The linear drift change of measure supplies risk-neutral dynamics $dS=rSdt+\sigma d{W}^{\mathbb{Q}}$, so the same price is the discounted conditional payoff expectation under the usual admissibility and integrability hypotheses.