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For the sphere model of the real projective plane, letThen ${\mathbb{RP}}^2=S^2/{\sim }_S$ with the quotient topology: each equivalence class is an antipodal pair.

For the closed disc model, letGive $D^2/{\sim }_D$ the quotient topology. Only antipodal pairs on the boundary circle are identified in the closed disc model; interior points remain distinct.

Let $W=(\mathbb{N}\times S^1)/A$, and choose in each circle the point opposite $o$:In the subspace topology from the Euclidean plane, the sequence $z_n$ converges to $o$, which does not belong to $B$. Thus $B$ is not closed in $X$.

On the other hand, let $D={\bar{f}}^{-1}(B)\subseteq W$. With the quotient map $q$ from part (i),This is closed in $\mathbb{N}\times S^1$: its complement is the open set $\mathbb{N}\times (S^1\setminus \{(0,1)\})$. The definition of the quotient topology, applied to complements, says that $D$ is closed in $W$. A homeomorphism would carry this closed set to the closed set $B$, a contradiction. Therefore $\bar{f}$ is not a homeomorphism.

The obstruction occurs at the common point. The open set $W\setminus D$ contains the collapsed point, but its image $X\setminus B$ contains no neighbourhood of $o$, since $z_n\to o$. The wedge sum topology allows a separate small neighbourhood on each circle; a neighbourhood of $o$ in the Hawaiian earring must contain all sufficiently small whole circles.

Use positive indices $n\ge 1$ and write $p=(0,-1)\in S^1$. SetFor fixed $n$, this function is a homeomorphism from the unit circle to $C_n$: it scales its radius to $1/n$ and translates its centre to $(0,1/n)$. Thus it takes values in the Hawaiian earring $X$ and is surjective onto $X$. It is continuous on each open slice $\{n\}\times S^1$; since these slices form an open cover of $\mathbb{N}\times S^1$, $f$ is continuous on the whole product topology.

Distinct circles $C_n,C_m$ meet only at $o=(0,0)$. To see this directly, their equations areFor $m\ne n$, subtraction gives $y=0$, then $x=0$. Each slice contains exactly one preimage of $o$, namely $(n,p)$. Therefore the only nonsingleton fibre of $f$ is $A$.

Let $q:\mathbb{N}\times S^1\to (\mathbb{N}\times S^1)/A$ be the quotient map collapsing $A$. The universal property of the quotient topology supplies a unique continuous map $\bar{f}$ with $f=\bar{f}\circ q$. The fibre calculation proves injectivity, and surjectivity follows from that of $f$. Hence $\bar{f}$ is the required continuous bijection.

Let $o=(0,0)$ be the common point of the two circles. For sufficiently small $\epsilon >0$, the open neighbourhoodhas the following local structure: $W\setminus \{o\}$ consists of four disjoint open arcs, an upper and a lower branch from each circle. These are its four connected components, each open and closed relative to $W\setminus \{o\}$.

Suppose a coordinate chart of some dimension of a manifold $n$ existed at $o$. The case $n=0$ would make $o$ an isolated point, which it is not. For $n\ge 1$, shrink the chart inside $W$ so that its image is an open ball centred at the image of $o$. Its domain of a function $V$ is an open neighbourhood of $o$, so it meets all four arms of $W$. Intersecting those arms with $V\setminus \{o\}$ gives four disjoint nonempty clopen sets partitioning $V\setminus \{o\}$; hence there are at least four connected components.

But a punctured ball in $\mathbb{R}$ has exactly two connected components, and a punctured ball in ${\mathbb{R}}^n$ for $n\ge 2$ is path-connected. A homeomorphism preserves connected components, giving a contradiction in every dimension of a manifold. Thus $X$ is not a topological manifold. The argument uses arbitrarily small coordinate charts; the fact that the whole $X\setminus \{o\}$ has two pieces alone would not be a local obstruction.

Label the hexagon polygon vertices $v_0,\dots ,v_5$ in the direction in which its boundary word is $abc{a}^{-1}{b}^{-1}{c}^{-1}$, with the first side running from $v_0$ to $v_1$. Matching the endpoints according to the arrows givesThus the two vertex equivalence classes are $\{v_0,v_2,v_4\}$ and $\{v_1,v_3,v_5\}$.

The polygonal schema gives a compact connected surface with empty manifold boundary. Interior points have disc neighbourhoods; a paired polygon edge glues two half-disc neighbourhoods into a disc. At each of the two vertex equivalence classes, its three corner sectors join cyclically, so the vertex link in a polygonal schema is a circle, again giving a disc neighbourhood. Pairing every label once positively and once negatively makes these gluings consistent with an orientation, so the resulting surface is orientable.

The resulting CW complex has CW cells comprising two vertices, three edges and one face. Its Euler characteristic isBy the classification of compact surfaces, a connected closed orientable surface with $\chi =0$ has genus $g$ satisfying $2-2g=0$, hence $g=1$. The quotient space is a torus.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

Let $g_1=x^2+y^2+z^2-4$ and $g_2=(x-3{)}^2+y^2+z^2-4$. Subtracting the two constraint equations gives $-6x+9=0$, hence $x=3/2$ and $y^2+z^2=7/4$. The feasible set is therefore a circle.

For Lagrange multipliers $\lambda ,\mu$, the stationarity equation $\nabla f=\lambda \nabla g_1+\mu \nabla g_2$ becomesAt $x=3/2$, the first equation gives $\lambda =\mu$. The second ensures $\lambda +\mu \ne 0$ and $y\ne 0$, so the third gives $z=0$. The constraints then force $y=\pm \sqrt{7}/2$. Both points do satisfy the multiplier equations, taking $\lambda =\mu =1/(4y)$.

There are no missing singular constraint points: on the feasible circle, the two constraint gradients $(3,2y,2z)$ and $(-3,2y,2z)$ are linearly independent, since $y^2+z^2=7/4>0$. Thus the complete list of constrained critical points and critical values isThe bound $|y|\le \sqrt{7}/2$ on the feasible circle shows that the positive value is the global maximum and the negative value is the global minimum.

Write $q(x,y)=x^2+y^2$. Equality of $q$-values defines an equivalence relation induced by a function: equality is reflexive, symmetric, and transitive, so all three properties hold for $R$. The equivalence class of $(x,y)$ isFor $(x,y)\ne (0,0)$ this is the circle centred at the origin with radius $\sqrt{x^2+y^2}$; the remaining equivalence class is the singleton $\{(0,0)\}$.

Periodicity is invariance of a function under a nonzero shift $P$. Its domain of a function must be preserved by the shift and its inverse, and $f(x+P)=f(x)$ at every allowed function input. On the real line, both $\sin x$ and $\cos x$ have period $2\pi$.

For an angular coordinate on a full circle, single-valuedness requires $2\pi$-periodicity. For nonzero real $k$, the functions $\sin (k\theta )$ and $\cos (k\theta )$ are $2\pi$-periodic precisely when $k$ is an integer: the phase must change by an integer multiple of $2\pi$. The zero-frequency case instead gives constant functions. Their Fourier series supply angular modes for the harmonic expansion on a disk.

For $L>0$, extend $f_L(x)=\sin x-x\sin L/L$ from $[-L,L]$ periodically. The linear term makes both endpoint values zero and the endpoint first derivatives match. The function is odd, so its Fourier series has only sine terms, with the displayed Fourier coefficients. Indeed, for $k_n=n\pi /L\ne 1$,subtracting and multiplying by $2/L$ proves the formula. The coefficients decay as $O(n^{-3})$. At $L=m\pi$, the single coefficient $b_m$ has removable limit $1$, with all others zero. Multiplying $b_n$ by $e^{-\kappa k_n^{2}t}$ solves the corresponding heat equation on the circle.

For pointed topological spaces $(X,x_0)$ and $(Y,y_0)$, the pointed mapping space ${\mathrm{Map}}_{\ast }(X,Y)$ consists of continuous maps $f:X\to Y$ with $f(x_0)=y_0$. It carries the subspace topology from the compact-open topology on the ordinary mapping space, commonly followed by kification when working with compactly generated spaces. Its basepoint is the constant map with value $y_0$.

For the pointed circle, ${\mathrm{Map}}_{\ast }(S^1,Y)$ is the loop space $\Omega Y$: a circle map is a path returning to its starting point. More generally, for well-behaved pointed spaces, path components of ${\mathrm{Map}}_{\ast }(S^n,Y)$ correspond to ${\pi }_n(Y,y_0)$. Thus a mapping space turns families of maps and homotopies into points and paths of another topological space. The choice of mapping-space topology ensures that this correspondence works continuously.

For $a>0$ and $0<c<2a$, consider the region $c\le r\le a(1+\cos \varphi )$ in polar coordinates. It exists precisely for $|\varphi |\le \beta$, where $\beta =\arccos (c/a-1)\in (0,\pi )$. The polar area element $rdrd\varphi$ givesExpanding the square and integrating its even terms gives the title formula. This cap lies inside the cardioid and outside the concentric circle, so reversing the radial bounds would select the wrong region or a signed area.

A conic section with a focus of a conic at the origin has the polar coordinates equation $r=\ell /(1+e\cos \theta )$, where $\ell >0$ is the semilatus rectum and $e\ge 0$ the eccentricity. The directrix of a conic definition $r=e(d-r\cos \theta )$ gives the formula with $\ell =ed$ when $e>0$; $e=0$ is the circle limit. The cases $0<e<1$, $e=1$, and $e>1$ give an ellipse, parabola, and hyperbola. The physical radius requires a positive denominator. For an attractive inverse-square central force, the Laplace-Runge-Lenz vector gives $\ell =m{h}^2/k$ and $e=A/(mk)$.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

For a polynomial $p(z)=c{\prod }_j(z-a_j)$ with $c\ne 0$, no polynomial root at zero, and $R>0$ different from every $|a_j|$, Jensen's formula gives the displayed expression. The polynomial roots are repeated according to their multiplicity of a zero: an exterior root contributes $\log |a_j|$, whereas an interior root contributes $\log R$.

For an explicit example, take the polynomialIts two simple zeros are $1/4$ and $-1/2$. The circle log-mean of a holomorphic function is thereforeAt $R=0.2$ the open disc contains no zeros of a function, at $R=0.4$ it contains $1/4$ only, and at $R=0.75$ it contains both. These three circles avoid the polynomial roots; the radii $1/4$ and $1/2$ are excluded from the displayed version.

The corresponding values of the circle log-mean of a holomorphic function and the circle geometric mean of a holomorphic function areFor example, at $R=0.4$ the contribution of the sole interior zero of a function is $\log (0.4/0.25)=\log (8/5)$; adding it to $\log (3/8)$ gives $\log (3/5)$. The second polynomial root still affects $f(0)$, but contributes no additional interior-zero term at this radius.

For a simple closed curve in the standard oriented complex plane, positive orientation keeps the bounded interior on the left while traversing the curve. A circle is therefore traversed counterclockwise; reversing direction changes the sign of its contour integral.

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.