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The map $h$ is a homeomorphism from the real line to its image with the subspace topology: its continuous inverse is $h^{-1}(x,y)=\log \sqrt{x^2+y^2}$. Different parameters have different radii, so winding does not destroy injectivity. The origin is an accumulation point of the spiral but is not a point of this space.

Let $A$ be a nonempty subset of a metric space $X$, and let $f:A\to \mathbb{R}$ satisfy a Lipschitz bound $L$. Thenare real-valued Lipschitz functions on all of $X$, agree with $f$ on $A$, and obey the same bound $L$. Every other extension $H$ satisfying that bound lies between them, $G\le H\le F$. Despite the infimum in its formula, $F$ is the largest such extension: every given sample supplies an upper constraint $H(x)\le f(a)+Ld(x,a)$.

Take three real samples $f(0)=0$, $f(1)=1/2$, $f(2)=0$, and allow the bound $L=1$. Their least Lipschitz constant is only $1/2$, so this is deliberately a looser permitted bound. The two extensions areAt $x=1/2$, the allowable interval is $0=G(1/2)\le H(1/2)\le F(1/2)=1/2$. The interpolant obtained by linear interpolation is a piecewise linear function and has value $1/4$ there. Extending it constantly outside $[0,2]$ obeys the same bound on the entire real line. At $x=5/2$, the allowable interval is $[-1/2,1/2]$. Thus the data and the bound control possible values without selecting a unique curve.

The construction is useful when data are known only on a subset: it extends them without increasing an allowed Lipschitz bound. A short justification explains why the envelopes work. Fix $a_0\in A$. The triangle inequality and the bound on $f$ implyso the infimum is finite below; the term $a=a_0$ makes it finite above. At a sample $x=a_0$, all upper terms are at least $f(a_0)$, while that term gives equality. Finally, $d(x,a)\le d(y,a)+d(x,y)$ gives $F(x)\le F(y)+Ld(x,y)$ after taking an infimum; exchanging $x,y$ proves the bound. The lower envelope follows by applying the same argument to $-f$.

E. J. McShane's paper Extension of range of functions appeared in the Bulletin of the American Mathematical Society, volume 40, in 1934, pp. 837–842. ^{[ref] [ref]}

The change of variables $x=e^u$, using the exponential function, turns the Mellin transform intoThe factor $dx=e^{u}du$ changes $x^{s-1}$ to $e^{su}$; missing this factor would shift every exponent by one. With $h_{\sigma }(u)=e^{\sigma u}f(e^u)$ and the Fourier transform convention $\hat{h}(\omega )={\int }_{\mathbb{R}}h(u)e^{-i\omega u}du$, the identity isChoosing $\sigma$ inside the Mellin fundamental strip makes $h_{\sigma }$ a Lebesgue integrable function. The minus sign in the Fourier transform argument follows from the stated convention. For the Mellin transform of an interval indicator with $[b,c]=[1,4]$ and $\sigma =0$, $h_0=1_{[0,\log 4]}$: the Fourier transform of an interval indicator is literally the same computation in a different coordinate.

Splitting the real line at $u=0$ also connects the Mellin transform to two unilateral Laplace transforms. If $k_{+}(v)=f(e^v)$ and $k_{-}(v)=f(e^{-v})$ for $v\ge 0$, thenwhere both integrals converge absolutely. Their Laplace convergence regions impose opposite endpoint conditions, explaining the two sides of the Mellin fundamental strip. Finally, $f(a{e}^u)=f(e^{u+\log a})$ shows that a change of scale is a translation in $u$. Weighted Fourier transforms of this translation give exactly the factor $a^{-s}$ in the scaling property of the Mellin transform. For a smooth function with compact support inside $(0,\infty )$, the Mellin inversion theorem follows from Fourier inversion; the existing formula states the corresponding hypotheses.

A nonconstant Möbius transformation mapping the real line into the unit circle has the displayed form with $|\lambda |=1$ and $k$ nonreal. Indeed $|ax+b{|}^2=|cx+d{|}^2$ for real $x$ yields $|a|=|c|\ne 0$, $|b/a|=|d/c|$ and $\mathrm{Re}(b/a)=\mathrm{Re}(d/c)$. Thus $d/c$ is $b/a$ or its complex conjugate; the first possibility makes the map constant, so the second is necessary. Conjugate numerator and denominator have equal modulus on the real line, proving the converse as well.

A real-valued function $f:D\to \mathbb{R}$, with domain of a function $D\subseteq \mathbb{R}$ in the usual order of the real numbers, is a monotonic function if it is either a nondecreasing function throughout $D$ or a nonincreasing function throughout $D$. The same direction must work for every pair $x<y$ in $D$. Being a strictly increasing function on one part and a strictly decreasing function on another does not make the whole function monotone: for example, $x^2$ on the real line changes direction at zero. A constant function is both a nondecreasing function and a nonincreasing function.

The word "increasing" is ambiguous across textbooks: some allow equal outputs and some require unequal outputs. Use nondecreasing function for the weak inequality and strictly increasing function for the strict inequality; use nonincreasing function and strictly decreasing function in the other direction. Neither continuity nor differentiability is part of the definition of a monotonic function.

Strict monotonicity gives injectivity, so an equation $f(x)=b$ has at most one solution and an inverse function exists on the image of a function. Weak monotonicity permits repeated outputs and jumps. In probability theory, a cumulative distribution function is a nondecreasing function; its quantile function uses a generalized inverse to handle those flat parts and jumps. Monotonicity alone does not guarantee continuity, even for a strictly increasing function.

A multiresolution analysis organizes a function into successively finer approximation spaces. In the usual orthonormal construction on $L^2(\mathbb{R})$, these are closed linear subspaces $(V_j{)}_{j\in \mathbb{Z}}$ withRescaling relates the levels: $f(x)\in V_j$ if and only if $f(2x)\in V_{j+1}$. Integer translates of a scaling function form an orthonormal basis of $V_0$. The detail space is the orthogonal complement of $V_j$ inside $V_{j+1}$, giving $V_{j+1}=V_j\oplus W_j$. Coarse information and new details therefore fit together without overlap in squared L2 norm.

For the Haar wavelet, $V_j$ consists of square-integrable functions constant on cells of length $2^{-j}$. Splitting every cell into two halves creates one new contrast per old cell, the Haar refinement identity. On the bounded example $[0,1]$, the first three spaces have dimensions $1,2,4$: a whole-interval mean, two half-interval means, and four quarter-interval means. The Haar projection diagram displays those three resolutions. This bounded construction starts at level zero; the displayed intersection condition refers to the full family on the real line.

The usefulness is a controlled way to increase resolution. Earlier coarse information remains available while new wavelet coefficients identify what the finer description adds. The wavelet approximation complexity connects this extra resolution to both approximation error and the number of coefficients being estimated. ^[ref]

A function is nonconstant when two inputs in its domain of a function have different outputs. For example, $f(x)=x^2$ on the real line is nonconstant, whereas the same rule restricted to $\{-1,1\}$ is a constant function.

A subset $A$ of a measurable space $(X,\Sigma )$ is non-measurable when $A\notin \Sigma$. On the real line, unless another sigma-algebra is specified, this commonly means that $A$ is not a Lebesgue measurable set. By completeness of Lebesgue measure, such a set cannot be contained in a Lebesgue null set.

For an example, use the axiom of choice to select one representative from each equivalence class of $[0,1]$ under $x\sim y$ if $x-y\in \mathbb{Q}$, and call the resulting set $V$. The translates $V+q$, for $q\in \mathbb{Q}\cap [-1,1]$, are pairwise disjoint, their union contains $[0,1]$, and their union is contained in $[-1,2]$. If $V$ were a Lebesgue measurable set, translation invariance and countable additivity would make that union have measure zero when $m(V)=0$ and infinite measure when $m(V)>0$. Both conclusions contradict these inclusions. Thus $V$ is non-measurable.

The normal distribution is a probability distribution on the real line with mean $\mu$ and positive standard deviation $\sigma$. Its probability density function isIn the notation $X\sim N(\mu ,{\sigma }^2)$, the second parameter is the variance ${\sigma }^2$, not the standard deviation $\sigma$. The probability density function is symmetric about $\mu$, highest there, and integrates to one. These formulas and the parameter convention are given in MIT 18.05, Reading 5c, §5: normal distributions.

The normal distribution is a location-scale family: if $Z$ is a standard normal random variable, then $X=\mu +\sigma Z$. Changing $\mu$ moves the center; changing $\sigma$ stretches the horizontal distances from it. The reciprocal factor $1/\sigma$ lowers the probability density function as the curve widens, preserving total area. This is why one standard normal distribution suffices to calculate probabilities for the whole family. ^[ref].

The height $f_X(x)$ is a probability density function value, not $\mathbb{P}(X=x)$. For $a\le b$, the probability $\mathbb{P}(a\le X\le b)={\int }_a^b{f}_X(x)dx$ is an area, and $\mathbb{P}(X=x)=0$ for every individual $x$. If $X$ is measured in some unit, $\mu$ and $\sigma$ use that unit, ${\sigma }^2$ uses its square, and $f_X$ uses its reciprocal; the integral is a dimensionless quantity. A change of measurement unit therefore changes the height without changing the probability. The area interpretation is developed in MIT 18.05, Reading 5b: continuous random variables.

The usefulness of the normal distribution goes beyond its simple shape. Under the central limit theorem, sums of independent and identically distributed random variables with finite positive variance, after centering and scaling, approach the standard normal distribution. Thus approximations by a normal distribution can arise even when the individual random variables have quite different probability distributions. The hypotheses matter: a model of measurement error still needs evidence that this approximation is appropriate. ^[ref]. Exact closure under a linear combination of independent normal random variables provides another reason the family is convenient.

Historically, the curve appeared in a different problem from fitting measurements. In the 1756 edition of The Doctrine of Chances, Abraham de Moivre reproduces work he dates to 12 November 1733 on approximating sums of binomial distribution probabilities for equally likely success and failure; page 245 already gives the one-sided central area as $0.341344$. ^[ref]. In his 1809 Theoria motus, Carl Friedrich Gauss derived a normal distribution for measurement error from a condition on the arithmetic mean, then connected maximizing its likelihood to minimizing the sum of squared errors under that model, the principle behind ordinary least squares. These are distinct contributions: approximating binomial distribution probabilities and justifying an estimation rule. ^[ref].

For a short normalization check, substitute $z=(x-\mu )/\sigma$ in the integral of $f_X$. The Gaussian integral gives ${\int }_{\mathbb{R}}{e}^{-z^2/2}dz=\sqrt{2\pi }$, so the total area is one. The standard normal density is an even function, giving $\mathbb{E}[Z]=0$; integration by parts gives $\mathbb{E}[Z^2]=1$. Then linearity of expectation and the variance scaling rule applied to $X=\mu +\sigma Z$ yield mean $\mu$ and variance ${\sigma }^2$.

On two copies of the real line with their usual topologies, let $U=(1,2)$ and $V=(2,4)$. The coordinate projection maps giveThe first two open sets are unbounded strips; their set intersection is an open rectangle. The point $(1.4,3)$ passes both coordinate tests, whereas $(2,3)$ fails the first because the endpoint $2$ is excluded.

Every point of an open set in the Euclidean plane has such an open rectangle around it inside that open set. Conversely, every open rectangle contains a small Euclidean ball around each of its points. Thus the product topology here is the usual metric topology of the Euclidean plane. With finitely many factors all coordinates may be restricted; on an infinite Cartesian product a basic cylinder set must leave all but finitely many factors free.

An open set is a member of the topology on a topological space. In a metric space, a subset $U$ is an open set precisely when every $x\in U$ has an open ball $B(x,r)$ contained in $U$ for some $r>0$. For $U=(0,1)$, one can take $r=\frac{1}{2}\min (x,1-x)$. At $x=0.2$, this gives $r=0.1$ and $B(x,r)=(0.1,0.3)\subset U$. The radius depends on the point; no uniform radius is required. ^[ref]

Openness depends on the ambient topological space. In $A=[0,1]$ with its subspace topology, $[0,1/2)=A\cap (-1,1/2)$ is an open set in $A$, although it is not an open set in the real line. Similarly, the whole closed interval $A$ is both an open set and a closed set in itself. These relative open sets are what the continuity test uses for maps involving a subset with its subspace topology.

Arbitrary intersections of open sets need not be open sets: ${\bigcap }_{n\ge 1}(-1/n,1/n)=\{0\}$, a singleton that is not an open set in the usual real line. Any point $x\ne 0$ is excluded once $n>1/|x|$. This is why the topology axiom specifies finite intersections.

The Pareto distribution describes a positive random variable whose tail probability falls as a power law. In the shape–minimum convention used here, $X\sim \mathrm{Pareto}(\alpha ,\beta )$ has shape parameter $\alpha >0$ and minimum $\beta >0$, with probability support $[\beta ,\infty )$. Its probability density function isThe minimum $\beta$ has the same units as $X$; the shape parameter $\alpha$ is dimensionless. At fixed $\alpha$, changing $\beta$ gives a scale family, since $X/\beta \sim \mathrm{Pareto}(\alpha ,1)$. This convention is also used in uniform-Pareto conjugacy.^[ref]

The survival function and cumulative distribution function areFor $0<p<1$, solving $F(x)=p$ gives the quantile functionIts right-hand limit at $p=0$ is $\beta$; it tends to infinity as $p\uparrow 1$.^[ref] There is no probability mass at $\beta$: changing a probability density function's value at that single point changes no probabilities.

Choose $\alpha =2$ and $\beta =10$ as an original theoretical example. Then $\mathbb{P}(X>20)=1/4$, $\mathbb{P}(X>100)=1/100$, the median is $10\sqrt{2}\simeq 14.142$, and the 0.99 quantile is $Q(0.99)=100$. These chosen values and all figures below describe mathematical probability distributions, not actual wealth observations.

On this log-log plot, write $r=x/\beta$ to obtain $\log \overline{F}(\beta r)=-\alpha \log r$. The slope is $-\alpha$, whereas the probability density function has power exponent $\alpha +1$. Multiplying a threshold by ten divides its tail probability by $10^{\alpha }$. A larger $\alpha$ therefore concentrates the Pareto distribution nearer $\beta$; a smaller $\alpha$ makes extreme values more plausible. This exact scaling is useful when modelling large positive losses or sizes over several orders of magnitude.

The Pareto distribution is a heavy-tailed distribution: for every $t>0$, the exponential moment $\mathbb{E}[e^{tX}]$ diverges because $e^{tx}{x}^{-\alpha -1}$ eventually grows without bound. In contrast, the Laplace distribution has exponentially decreasing tails and some finite positive exponential moments. The Student's t-distribution also has power tails, but on both sides of the real line, with a smooth central region rather than a positive minimum.

For rank-one two-isogeny descent with parameters three and minus three, use the standard two-isogeny rank formula. For $C:y^2=x(x^2+ax+b)$ and its dual curve $D:y^2=x(x^2-2ax+a^2-4b)$, let ${\alpha }_C,{\alpha }_D$ be the abscissa square-class maps, with their usual values at $(0,0)$ and the identity. If $r_C$ is the Mordell–Weil rank, thenFor the present curve $a=3,b=-3$, its possible classes are $1,-1,3,-3$, and the identity and $(0,0)$ supply $1,-3$. Exclude the class three as follows. A point in that class has $x=3(U/V{)}^2$, with coprime integers $U,V$, and substitution in the curve gives the necessary quartic equationwith integer $N$. Modulo three this forces $3\mid V,N$. Coprimality then makes $U$ a unit modulo three, and modulo nine the right side is $3U^4\equiv 3$, while $N^2\equiv 0$, a contradiction. The class minus one is excluded as well, since multiplying it by the known class minus three would give three. Therefore the first image has size two.

The dual curve is $D:y^2=x(x^2-6x+21)$. Its quadratic factor is $(x-3{)}^2+12>0$ on the real line, so every nonzero rational abscissa is positive. By (i) its image is contained in $\{1,3,7,21\}$. The point $(0,0)$ supplies 21 and the point $(3,6)$ supplies three. As the square-class map is a homomorphism, their product supplies seven. All four occur. Hence the descent formula gives $2^{r_C+2}=2\cdot 4=8$, so

The lower limit topology on the real line is the lower limit topology. Every nonempty basic interval $[a,b)$ contains a rational number, so the rational numbers are a countable dense subset. By the preceding product argument, ${\mathbb{Q}}^2$ is dense in the square, proving its topological separability.

Consider the uncountable anti-diagonalIt is a discrete subset: the basic rectangle $[t,t+ϵ)\times [-t,-t+ϵ)$ intersects $D$ only at $(t,-t)$, because both coordinates have increased weakly and their sum can remain zero only if neither increases.

It is also closed. At a point $(u,v)$ off the anti-diagonal, $u+v\ne 0$. If this sum is positive, every sufficiently small lower-limit rectangle based at $(u,v)$ still has positive coordinate sum. If it is negative, choose $ϵ>0$ with $2ϵ<-(u+v)$; then $[u,u+ϵ)\times [v,v+ϵ)$ has negative coordinate sum throughout. These neighborhoods avoid $D$, so its complement is open.

If the square were Lindelöf, its closed discrete subset $D$ would be countable by part (c), contradicting its bijection with $\mathbb{R}$. Hence

Fix a horizontal line $L=\mathbb{R}\times \{c\}$. The restriction of every member of $\mathcal{H}$ to $L$ is continuous in its usual topology. Therefore every subbasic open set intersects $L$ in a usual open set, so the induced topology is no finer than the usual one. Conversely the first coordinate projection is continuous for the initial topology, and its inverse images induce every usual open set on $L$. The two topologies on $L$ are equal. The second projection gives the identical argument for a vertical line.

Write $F={\overline{Y}}^X$. For every nonzero rational $r$, the intersection of $Y$ with $\{r\}\times \mathbb{R}$ is dense in that vertical line, since the nonzero rational numbers are dense in the real line. Hence $\{r\}\times \mathbb{R}\subseteq F$. For each real $c$, it follows that $F\cap (\mathbb{R}\times \{c\})$ contains $(\mathbb{Q}\setminus \{0\})\times \{c\}$ and is closed in that line. This dense subset forces the whole horizontal line into $F$. Thus $F=X$: $Y$ is a countable dense subset. The two successive uses of line density are necessary; the initial topology is finer than the Euclidean topology, so Euclidean density alone would not suffice.

Let $\mathcal{H}$ be the set of separately continuous real-valued functions on the plane. Take as a subbasis all sets $h^{-1}(U)$, where $h\in \mathcal{H}$ and $U$ is open in the usual real line. Unions of finite intersections of these sets form a topology, and every $h$ is continuous for it by construction. Conversely every topology making every $h$ continuous must contain the subbasis, hence this entire generated topology. Thus the required coarsest topology exists and is unique: it is the initial topology of the family $\mathcal{H}$.

The family is a set, being a subset of ${\mathbb{R}}^{{\mathbb{R}}^2}$, so this construction has no proper-class issue. The coordinate projections belong to $\mathcal{H}$; consequently this topology contains the ordinary Euclidean topology. This last observation will also make convergence in the new topology imply ordinary coordinatewise convergence.

Suppose for contradiction that $p_n=(u_n,v_n)\in Y$ converges to the origin in the initial topology. The coordinate projections then give $u_n\to 0$ and $v_n\to 0$ in the usual real line. Rectangular bumps detecting nonconvergence, using anisotropic supports, avoids any unsupported regularity assertion about a radial maximum. SetThe radii are positive since neither coordinate of $p_n$ vanishes. Each rectangular support misses both coordinate axes, so $B$ vanishes on those axes. For a fixed $y=c\ne 0$, all but finitely many vertical support intervals miss $c$, because their centres and radii tend to zero. Hence $B(x,c)$ is a finite maximum of continuous functions of $x$, and is continuous. At $c=0$ it is identically zero. The same argument with the coordinates exchanged proves separate continuity everywhere. Thus $B$ is continuous for the initial topology.

However $B(p_n)=1$ for every $n$, whereas $B(0,0)=0$. This contradicts preservation of convergent sequences by a continuous function. Therefore no sequence from $Y$ converges to the origin.

The radial hint in the scan is not separately continuous for every candidate sequence: for $p_n=(2^{-n},2^{-2n})$, its value at the origin is zero, while its value at $(2^{-n},0)$ is at least $1-2(2^{-2n})/\sqrt{2^{-2n}+2^{-4n}}$, which tends to one. The rectangular construction supplies the needed valid proof independently of that hint.

Let $h:Y\to \mathbb{R}$ be continuous, without a boundedness assumption. The homeomorphism $\theta (t)=t/(1+|t|)$ maps the real line onto $(-1,1)$, with inverse ${\theta }^{-1}(s)=s/(1-|s|)$. The bounded function $g=\theta \circ h$ extends by the preceding bounded construction to a continuous real-valued $G$ on $X$. Since $|G|<1$ on $Y$, the strict-range modification gives a continuous $\tilde{G}$ with range in $(-1,1)$ and the same restriction to $Y$. Thereforeis continuous everywhere: its denominator is pointwise positive. On $Y$ it equals ${\theta }^{-1}(\theta (h))=h$. Thus $Y$ is $C$-embedded. It is important to modify the extension before applying the inverse: an extension allowed to attain $1$ or $-1$ outside $Y$ would make this formula undefined there.

Write $T_n=h(\{x\in \mathbb{R}:|x|>n\})$, making explicit that these are subsets of $Y$. If $p\notin h[\mathbb{R}]$ and $U$ is an open neighbourhood of $p$, then $h([-n,n])$ is compact and hence closed in $Y$. The open set $U\setminus h([-n,n])$ still contains $p$ and meets the dense image of $h$. Such an intersection point belongs to $T_n$, so $p\in \overline{T_n}$ for every $n$.

Conversely, if $p=h(x)$ choose an integer $n>|x|$. Since $h$ is an embedding, there is an open neighbourhood $U$ of $p$ in $Y$ whose intersection with $h[\mathbb{R}]$ is $h((-n,n))$. It misses $T_n$, so $p\notin \overline{T_n}$. HenceFor the two ends of a real-line compactification, let $A_n^{+}=\overline{h((n,\infty ))}$ and $A_n^{-}=\overline{h((-\infty ,-n))}$. Each is nonempty compact and connected, being the closure of a connected image. The two nested families therefore have nonempty connected intersections $C^{+}$ and $C^{-}$. Since closure commutes with finite unions, $\overline{T_n}=A_n^{+}\cup A_n^{-}$. MoreoverFor if a point fails both intersections, it fails one positive member and one negative member, hence both members at the larger index by nesting.

If the remainder were a finite three-point set, its Hausdorff subspace topology would be discrete. Each nonempty connected subset $C^{\pm }$ would then be a singleton, so their union would have at most two points. This contradiction proves there is no three-point Hausdorff compactification of the real line. The argument does not assert that every possible remainder is finite.

For $x<y$, finite additivity on the disjoint pieces of the half-line givesBy monotonicity of a measure and the Lebesgue measure of an interval,Exchanging the two points if necessary provesThus the cumulative measure of a finite-measure set is a Lipschitz function with Lipschitz constant at most one, hence is uniformly continuous: given $\epsilon >0$, the choice $\delta =\epsilon$ works simultaneously at every point.

To locate its endpoint limits, the measurable sets $E\cap (-\infty ,N]$ increase to $E$, so continuity from below of a measure gives $h(N)\to m(E)$. The measurable sets $E\cap (-\infty ,-N]$ decrease to the empty set and all have finite measure, so continuity from above of a measure gives $h(-N)\to 0$. Since $h$ is nondecreasing,For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The intermediate value theorem on $[-R,R]$ gives $t$ with $h(t)=c$. ThenThe endpoint values $c=0,m(E)$ can also be achieved by $A=\varnothing ,E$. This proves divisibility of Lebesgue measure for arbitrary finite-measure Lebesgue measurable sets on the real line, including unbounded ones. The non-atomic measure structure permits prescribed fractions of measure; a measure concentrated at one point would not have this property.