A metric space is path-connected if every pair $x,y$ can be joined by a path: a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.The path must remain in $X$, including at its endpoints.

A metric space is connected when it has no separation of a topological space: it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, the only subsets that are both open and closed are the empty set and the whole space.The open sets here are those of the metric topology.

The points $(n,0)$ belong to the indicated subset for every positive integer $n$. They form an unbounded sequence, so this metric space is not bounded. Therefore it is not compact. Shrinking the vertical width does not bound its horizontal extent.

The open unit disk is bounded but not closed: $(1-1/n,0)$ lies in it for $n\ge 2$ and converges to the omitted point $(1,0)$. By the Heine-Borel theorem, it is not compact.

The unit circle is the inverse image of the closed set $\{1\}$ under the continuous map $(x,y)\mapsto x^2+y^2$, so it is closed. It is also bounded, since every point has Euclidean norm one. The Heine-Borel theorem gives compactness.

Take an irrational number $a\in (0,1)$. The density of the rational numbers provides a sequence in $\mathbb{Q}\cap [0,1]$ converging to $a$, which is outside that subset. Thus the subset is not closed in $\mathbb{R}$. By the Heine-Borel theorem, it is not compact, despite being bounded.

The sequence $n\in [0,\infty )$ is unbounded. Every compact metric space is bounded, by part (b), so this subspace is not compact.

Choose a Möbius map $T$ with $T(z_1)=0$ and $T(z_2)=\infty$, for example by normalizing the ordered triple $(z_1,z_3,z_2)$. Conjugation $h\mapsto T\circ h\circ T^{-1}$ takes the requested subgroup to the subgroup fixing zero and infinity individually.

Such a transformation must be $z\mapsto cz$ with $c\in {\mathbb{C}}^{\ast }$: the zero condition removes the constant numerator term and the infinity condition removes the nonconstant denominator term. Conversely every such dilation is Möbius and fixes the two points. Composition multiplies its nonzero parameter.This identifies the Möbius pointwise stabilizer of two points. Allowing an interchange of the two fixed points would give a larger group, but those interchanges are excluded by the stated equalities.

Let $G$ be the subgroup preserving the three-point set. Its action gives a group homomorphism $\Phi :G\to S_3$. If $\Phi (h)$ is the identity permutation, then $h$ fixes all three points, so part (i) forces $h$ to be the identity Möbius map. Thus the kernel is trivial and $\Phi$ is injective.

For any permutation $\sigma$, part (i) gives a normalization $g_{\sigma }$ sending the ordered triple $(z_{\sigma (1)},z_{\sigma (2)},z_{\sigma (3)})$ to $(0,1,\infty )$. If $g$ normalizes the original triple, then $h_{\sigma }=g_{\sigma }^{-1}\circ g$ sends $z_j$ to $z_{\sigma (j)}$. Hence every permutation is attained, so $\Phi$ is surjective.This is the setwise stabilizer; its six maps need not fix each point individually.

When all three points are finite, useThe distinctness assumptions ensure that it is a nonconstant Möbius transformation, and direct substitution gives the three required values. If one of the specified points is infinity, the limiting formulas areThese cover every case because the points are distinct.

For uniqueness, if $g$ and $\tilde{g}$ have the same three images, the Möbius map $\tilde{g}\circ g^{-1}$ fixes $0,1,\infty$. Fixing zero and infinity makes it $z\mapsto cz$ with $c\ne 0$; fixing one gives $c=1$. Therefore the normalized map is unique. This is the sharp three-transitivity of Möbius transformations.

For $z=x+iy$ and $r^2=|z{|}^2$, solving the sphere and projection equations givesFor $z\ne 0$, the reciprocal has real and imaginary parts $x/r^2$ and $-y/r^2$, and squared modulus $1/r^2$. Substituting these into the inverse formula producesThusThe reciprocal exchanges zero and infinity, corresponding to exchanging the south and north poles, so the formula holds there too. On the sphere this is a rotation through $\pi$ about the first coordinate axis, not the antipodal map.

The line through the north pole and $p=(a,b,c)$ isIts intersection with the equatorial complex plane has third coordinate zero, giving $t=1/(1-c)$ when $p$ is not the north pole. The first two coordinates are therefore $a/(1-c)$ and $b/(1-c)$, henceThe north pole maps to the point at infinity. This is stereographic projection onto the Riemann sphere.

Let the common domain be an open set $U$. The uniform limit theorem makes the limit $f$ continuous. For every triangle whose closed interior lies in $U$, Cauchy's integral theorem gives ${\int }_{\partial T}{f}_n=0$. Part (ii) passes this identity to the limit, giving ${\int }_{\partial T}f=0$. Morera's theorem now proves that $f$ is holomorphic.

The same reasoning works for locally uniform convergence, since each triangle boundary is compact. No passage of derivatives through a limit has been assumed.

For the usual piecewise smooth complex path $\gamma$, let $\ell (\gamma )$ denote its finite length. The estimation lemma givesBy uniform convergence, the supremum tends to zero, soThe uniform limit is continuous on the path, so its integral exists. This is uniform convergence and contour integration; the path length is fixed while the functions vary.

Morera's theorem says that a continuous complex-valued function $f$ on an open set $U\subseteq \mathbb{C}$ is holomorphic iffor every triangle $T$ whose closed interior is contained in $U$. Triangle boundaries are traversed with positive orientation. The continuity hypothesis matters: the zero-integral condition is combined with continuity, not used in isolation.

The map $h(z)=e^{\pi z/2}$ is a conformal bijection from the width-two strip to the upper half-plane, and $h(i)=i$. Its inverse there is $h^{-1}(w)=2L(w)/\pi$.

Write $e^{\pi \beta /2}=A+iB$, where $B>0$. Conjugate the preceding affine half-plane map by $h$:Every argument of $L$ lies in the upper half-plane. The formula is therefore a conformal bijection of the strip onto itself, andThe final equality uses $0<\mathrm{Im}(\pi \beta /2)<\pi$, so no multiple of $2\pi i$ is lost. This is a strip automorphism with a prescribed image of its midpoint.

Write $\alpha =a+ib$ with $b>0$. ThenIt preserves the upper half-plane, has inverse $(z-a)/b$, and sends $i$ to $\alpha$. Its nonzero derivative $b$ verifies that this bijection is conformal.

In the upper half-plane, the argument lies strictly between zero and $\pi$, while $\log |z|$ can be any real number. Conversely, if $w=s+it$ with $0<t<\pi$, then $z=e^w$ is in the upper half-plane and $L(z)=w$.Thus the principal complex logarithm is a conformal bijection from the upper half-plane onto that horizontal strip, with inverse the exponential function restricted to the strip.

A holomorphic primitive on the slit domain is $P(z)=zL(z)-z$, since the product rule and $L^{\prime }=1/z$ give $P^{\prime }=L$. The line segment stays in that domain. The fundamental theorem for line integrals therefore givesHenceThe value $L(i)=\pi i/2$ is fixed by the chosen branch.

Use the principal complex logarithm:The excluded nonpositive real ray allows a continuous choice of argument, and $L(1)=0$. For $z=x+iy$, its real part is $u=\frac{1}{2}\log (x^2+y^2)$. Differentiating the chosen local argument givesThese continuous partial derivatives satisfy the Cauchy-Riemann equations, proving holomorphy on the slit domain. Its derivative is