Yes. Let $(x^{(k)}{)}_{k\ge 1}$ be any sequence in $A$, where $x^{(k)}=(x_m^{(k)}{)}_{m\ge 1}$. At coordinate $m$, the values lie in the compact interval $[-1/m,1/m]$. By part (a)(ii) we can successively extract a convergent subsequence for coordinate one, a further one for coordinate two, and so on. Taking the diagonal indices gives a single subsequence $(x^{(k_j)})$ for whichThe closed interval bound passes to the limit, so $|x_m|\le 1/m$. Hence $x_m\to 0$ and $x=(x_m)\in A$.

Coordinate convergence alone would not imply supremum norm convergence. Here the common tail bound supplies the missing estimate. Given $ϵ>0$, choose $M$ with $2/(M+1)<ϵ$. For $m>M$,For the finitely many $m\le M$, coordinate convergence makes all differences less than $ϵ$ once $j$ is large. Therefore $d(x^{(k_j)},x)<ϵ$, andThis is an instance of compactness from uniformly vanishing sequence tails. The common $1/m$ bound is exactly what the coordinate vectors in part (i) lack.

The source prints $m\ne 0$ in the zero case. Since the coordinates are indexed by positive integers, that overlaps its $m=n$ case and does not define a sequence. We use the intended coordinate vectors, with $e_m^{(n)}=1$ for $m=n$ and $e_m^{(n)}=0$ for $m\ne n$.

Every $e^{(n)}$ belongs to $X$, because its entries are bounded by one and its coordinate sequence eventually vanishes. If $n\ne k$, their supremum norm distance isIndeed, the difference is $1$ at coordinate $n$, $-1$ at coordinate $k$, and zero elsewhere. No subsequence is a Cauchy sequence, since arbitrarily late distinct terms remain distance one apart. Every convergent sequence is a Cauchy sequence, so no convergent subsequence exists. Therefore $\boxed{X\text{ is not sequentially compact}}$.

Let $(x_n)$ be any sequence in the compact metric space $X$. There must be an $x\in X$ such that every open ball about $x$ contains $x_n$ for infinitely many indices $n$. Otherwise, for each $x\in X$ choose a ball $B(x,r_x)$ containing only finitely many terms, counted by indices. These balls are an open cover. A finite subcover would imply that only finitely many indices occur anywhere in $X$, a contradiction.

Now choose successively $n_k>n_{k-1}$ with $x_{n_k}\in B(x,1/k)$; infinitely many available indices make this possible. Then $d(x_{n_k},x)<1/k\to 0$, so $(x_{n_k})$ is a convergent subsequence. ThusThe proof counts indices rather than distinct points, so it also handles a value repeated infinitely often.

A metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ cover $X$.

It is a sequentially compact space if every sequence $(x_n)$ in $X$ has a convergent subsequence $(x_{n_k})$ whose limit of a sequence belongs to $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

Let $(F_n)$ be a Cauchy sequence for the Hausdorff distance, and putThe bounded set hypothesis makes ${\delta }_n$ finite, and the Cauchy sequence property gives ${\delta }_n\to 0$. Each $A_n$ is nonempty and a closed set, with $A_{n+1}\subseteq A_n$.

We need estimates that survive taking the closure. For every nonempty $F$, the distance to a set is Lipschitz continuous with constant one: the triangle inequality gives $\mathrm{dist}(x,F)\le d(x,y)+\mathrm{dist}(y,F)$, and interchanging $x,y$ provesFor every $m\ge n$, points of $F_m$ have distance at most ${\delta }_n$ from $F_n$. The same holds throughout their union and its closure, by this continuity. Since $F_n\subseteq A_n$,More generally, if $m\ge n$, every point in ${\bigcup }_{k\ge n}{F}_k$ is within ${\delta }_n$ of $F_m\subseteq A_m$. Passing to its closure again, and using $A_m\subseteq A_n$, givesChoose strictly increasing indices $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. Then $(A_{n_j})$ satisfies part (ii), so it converges in Hausdorff distance to the nonempty closed set $G={\bigcap }_j{A}_{n_j}={\bigcap }_n{A}_n$, with $\rho (A_{n_j},G)\le 2^{1-j}$.

For $n\ge n_j$, the triangle inequality gives the explicit full-sequence boundIt tends to zero as $j\to \infty$. ThereforeThis proves Completeness of the Hausdorff hyperspace using the proposed tail closures, rather than assuming their diameters shrink.

The intersection $G$ is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.

Fix $n$, $x_n\in F_n$, and $\eta >0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ is a Cauchy sequence. By completeness it converges to some $x\in X$.

For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne \varnothing$. Moreover,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.

Write $h(A,B)={\sup }_{a\in A}\mathrm{dist}(a,B)$, so that $\rho (A,B)=\max \{h(A,B),h(B,A)\}$ is the Hausdorff distance. Since $X$ is a bounded set, there is a finite $M$ with $d(x,y)\le M$. Nonemptiness of $A,B$ therefore gives $0\le h(A,B)\le M$, so $\rho$ is finite. Its symmetry and nonnegativity are immediate, as is $\rho (A,A)=0$.

If $\rho (A,B)=0$, then every $a\in A$ has zero distance to a set $B$. Since $B$ is a closed set, part (a)(iii) gives $A\subseteq B$. Interchanging the sets gives $B\subseteq A$. Thus $\rho (A,B)=0$ implies $A=B$. Closedness is essential here: without it a set and its closure have Hausdorff distance zero.

For the triangle inequality, fix $a\in A$ and $ϵ>0$. The approximation property of an infimum supplies $b\in B$ with $d(a,b)<\mathrm{dist}(a,B)+ϵ$, and then $c\in C$ with $d(b,c)<\mathrm{dist}(b,C)+ϵ$. Consequently,Let $ϵ\downarrow 0$ and take the supremum over $a$ to obtain $h(A,C)\le h(A,B)+h(B,C)$. Applying this inequality in the reverse direction as well yieldsAll metric axioms follow. Boundedness was used to ensure finiteness, and closedness to distinguish different sets; neither was needed for the triangle inequality.

First suppose $F$ is a closed set. If $x\notin F$, the definition in part (ii) gives an open ball $B(x,r)$ disjoint from $F$. Thus every $y\in F$ has $d(x,y)\ge r$, and the distance to a set satisfies $\mathrm{dist}(x,F)\ge r>0$. Its contrapositive is the required implication.

Conversely, suppose that zero distance to a set forces membership. For $x\notin F$ we have $\mathrm{dist}(x,F)>0$. Choose $0<r<\mathrm{dist}(x,F)$. Then $B(x,r)\cap F=\varnothing$, proving $X\setminus F$ is an open set. HenceFor $F=\varnothing$, use the extended-real convention $\inf \varnothing =+\infty$: both assertions hold, with the implication vacuous. Part (b) only uses nonempty sets, so all its distances to a set are ordinary finite real numbers.

A subset $F\subseteq X$ is a closed set in $X$ when its complement is an open set in the metric topology. Equivalently,The open ball is taken inside the ambient metric space $X$; closedness always depends on that ambient space.

A metric space $(X,d)$ is a bounded set if there is a finite constant $M\ge 0$ such thatFor nonempty $X$, this is equivalent to containment in an open ball of finite radius: the triangle inequality converts a radius bound into a diameter bound, and conversely a diameter bound gives a ball of any slightly larger radius about a chosen point.

It is a complete metric space if every Cauchy sequence in $X$ has a limit of a sequence belonging to $X$. Explicitly, the hypothesis on $(x_n)$ isand completeness requires an $x\in X$ with $d(x_n,x)\to 0$. ThusBoundedness controls distances; completeness ensures that these limiting points remain in the space.

Use the same principal square root $S(z)=\sqrt{(z-a)/(z-b)}$, which is holomorphic on $A$ because its argument avoids the nonpositive real axis. Thenis a holomorphic function satisfyingThus it is a holomorphic branch of the requested square root. This construction does not require a global holomorphic logarithm of $z-a$ or $z-b$ separately. Its sign is fixed by $S(z)\to 1$ at infinity, so $g(z)/z\to 1$. The negative of this function is the other holomorphic branch; any two branches have a continuous quotient taking values in $\{1,-1\}$, which is constant because $A$ is connected.

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

The prescribed pole lets us choose $c=4$ and $d=-5-3i$. Write $f(z)=(Az+B)/(4z-5-3i)$. Fixing $1/2$ and $2$ givesso $A=5-3i$ and $B=-4$. ThusIts determinant is $-18\ne 0$, so it is a genuine Möbius transformation. Direct substitution gives $f(1/2)=1/2$, $f(2)=2$, and a pole at $(5+3i)/4$; the numerator at that pole is $9/2\ne 0$.

To identify the image of the unit disc, putThen $|\alpha {|}^2=8/17<1$, $|\lambda |=1$, andIn particular, its pole is outside the closed unit disc. The identityshows $f(D)\subseteq D$. For any $|w|<1$, solve $w=\lambda (z-\alpha )/(1-\overline{\alpha }z)$ to getIts denominator is nonzero, and the same modulus identity for this inverse gives $|z|<1$. Thus $f(D)=D$, and global injectivity of a Möbius transformation proves the required one-to-one correspondence.

Write $\Delta =ad-bc>0$. For real $a,b,c,d$ and $z\in H$, the denominator cannot vanish: if $c\ne 0$ its zero $-d/c$ is real, and if $c=0$ then $d\ne 0$. Taking the imaginary part after multiplying by $\overline{cz+d}$ givesThus $f(H)\subseteq H$. Alsoso it is conformal. Cross-multiplication shows that $f(z_1)=f(z_2)$ implies $\Delta (z_1-z_2)=0$, proving injectivity.

The inverse Möbius transformation isIts coefficients are real and its determinant is again $\Delta >0$, so the same imaginary part identity shows that it maps $H$ into $H$. Therefore every $w\in H$ has a preimage in $H$, proving surjectivity. Hence $f$ maps $H$ one-to-one and onto itself.

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

The Riemann mapping theorem gives a conformal bijection $G\to D$ exactly whengiven the hypothesis that $G$ is a simply connected domain. A domain is nonempty, open, and connected, so no extra nonemptiness qualification is needed.

For necessity, a conformal bijection $\mathbb{C}\to D$ would be a bounded nonconstant entire function, contradicting the Liouville theorem. For sufficiency, $G$ is a proper simply connected domain, exactly the hypothesis of the Riemann mapping theorem.

Suppose $w\in \mathbb{C}$ were omitted by $g$. Thenwould be an entire function. The reverse triangle inequality gives $|g(z)-w|\ge |g(z)|-|w|\to \infty$ as $|z|\to \infty$, so $F(z)\to 0$ there. On a sufficiently large closed disk, $F$ is bounded by continuity and the extreme value theorem; outside it, the limit gives a bound. The Liouville theorem therefore makes $F$ constant, and the limit forces $F\equiv 0$.

That is impossible, since $1/(g(z)-w)$ is nonzero at every finite $z$. No value $w$ can be omitted, and hence

The Liouville theorem states that every bounded entire function is constant.

By the Laurent theorem, $f$ has a unique Laurent series $f(z)={\sum }_{n=-\infty }^{\infty }{c}_n{z}^n$ on $R<|z|<S$. Split it intoHere $g$ is holomorphic on $D_S$ and $h$ is holomorphic on $A_{R,\infty }$. To check their extended regions of convergence explicitly, fix $r\in (R,S)$ and put $M_r={\max }_{|z|=r}|f(z)|$. The Laurent coefficient formula givesThe first bound ensures convergence uniformly on every closed disk $|z|\le r^{\prime }<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\ge r^{\prime \prime }>r$; letting $r$ range down to $R$ covers $A_{R,\infty }$. Locally uniform convergence of these power series, in $z$ and $1/z$ respectively, proves the claimed holomorphy. On the original annulus their sum is $f$.

The second estimate also proves $h(z)\to 0$ as $|z|\to \infty$: for any fixed $r\in (R,S)$ and $|z|>r$,

For uniqueness under this condition, suppose $f=g_1+h_1=g_2+h_2$. Define $F=g_1-g_2$ on $D_S$ and $F=h_2-h_1$ on $A_{R,\infty }$. The two definitions agree on the overlap, and their domains cover $\mathbb{C}$, so they define an entire function. Moreover, $F(z)\to 0$ at infinity. It is bounded outside a large disk by this limit, and inside by continuity and the extreme value theorem. The Liouville theorem makes $F$ constant, and its limit makes that constant zero. Hence $g_1=g_2$ and $h_1=h_2$. Without the limiting condition, adding a constant to $g$ and subtracting it from $h$ would preserve the decomposition.

Start with the partial fraction decompositionFor $1<|z|<2$, expand the first fraction in negative powers and the second in nonnegative powers using the geometric series:Thus the Laurent expansion on this annulus isFor $|z|>2$, both fractions expand in negative powers:Combining the two geometric series givesThe $z^{-1}$ coefficient is zero, consistently with the original rational function being $O(z^{-2})$ at infinity. Each displayed series converges absolutely on its stated annulus and uniformly on every compact subset of that annulus.

For $z\ne 0$, the denominator vanishes exactly when $1/z=(2k+1)\pi i$. Thus setEach denominator zero is simple, sinceEvery $z_k$ is purely imaginary and nonzero. The identity $\sin (iy)=i\sinh y$ for real $y$ shows $\sin z_k\ne 0$, so there is no cancellation. Each $z_k$ is consequently a simple pole, with residue $z_k^2\sin z_k$.

These poles accumulate at zero. ThereforeIn particular, zero has no punctured neighbourhood on which the function is holomorphic, so it cannot be classified as a removable singularity, a pole, or an isolated essential singularity. There are no other finite singularities.