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A metric space is complete if every Cauchy sequence in it converges to a point of that space. The stated shrinking-set characterization is the Cantor intersection theorem together with its converse.

First assume completeness, and choose $x_n\in A_n$. If $m,n\ge N$, then both points belong to $A_N$, so $d(x_m,x_n)\le \mathrm{diam}(A_N)$. The vanishing diameters imply that $(x_n)$ is a Cauchy sequence, hence it converges to some $x\in X$. For every fixed $N$, its tail belongs to the closed set $A_N$, so $x\in A_N$. Thus $x\in {\bigcap }_N{A}_N$. In fact the intersection contains only one point, since any two of its points have distance at most every diameter.

Conversely assume the stated intersection property and let $(x_n)$ be a Cauchy sequence. PutThese closures are nonempty, closed and nested. Taking a closure does not increase the diameter: approximate two points of the closure by points of the set and use continuity of distance. The Cauchy sequence therefore makes $\mathrm{diam}(A_N)\to 0$. Choose $x$ in the intersection. For $n\ge N$, both $x_n$ and $x$ belong to $A_N$, and $d(x_n,x)\le \mathrm{diam}(A_N)\to 0$. Every Cauchy sequence converges in $X$, proving completeness.

We argue by contradiction using the previous subsequence construction. If $X$ is not complete, choose a Cauchy sequence with no limit in $X$. None of its $d_m$ can be zero: if $d_m=0$, then $d(x_m,x_n)\to 0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ withLet $Y=\{y_1,y_2,\dots \}$. This subset is closed in $X$. Indeed, if a point $x\notin Y$ lay in its closure, there would be points of $Y$ tending to $x$. Their indices must be unbounded, because the distance from $x$ to any finite subset of $Y$ is positive. Passing to increasing indices gives a convergent subsequence of the original Cauchy sequence. Part (a) would then make the whole sequence converge to $x$, a contradiction.

Now define $F:Y\to Y$ by $F(y_j)=y_{j+1}$. Distinctness makes this well-defined, and the displayed inequality makes it a contraction mapping with constant at most $1/2$. It has no fixed point, contradicting the assumed property of closed subsets. ConsequentlyThis is the converse direction of contraction property of closed subsets characterizes completeness.

The convention in this question counts a point whenever every open neighbourhood meets $A$, without excluding the point itself. These are adherent points, whose set is the closure $\overline{A}$; they differ from limit points under the convention that requires a different point of $A$ in each neighbourhood.

If $A$ is a closed set and $x\notin A$, the open set $X\setminus A$ is a neighbourhood of $x$ disjoint from $A$. Thus every adherent point lies in $A$. Conversely, if every adherent point lies in $A$, each $x\notin A$ has an open neighbourhood disjoint from $A$. Their union is $X\setminus A$, which is open. Hence $A$ is closed exactly when it contains all the points specified in the question.

The interior of a set and closure areThe neighbourhood description of the closure follows because failure to be adherent is equivalent to lying in an open set disjoint from $A$.

For closure of a connected set, suppose $\overline{A}=U\bigsqcup V$ were a separation into two nonempty relatively open sets. Their intersections with the connected subset $A$ would separate $A$ unless one intersection were empty. Say $A\subseteq U$. Choose $v\in V$ and an open set $O\subseteq X$ with $v\in O$ and $O\cap \overline{A}\subseteq V$. Since $v\in \overline{A}$, the set $O$ meets $A$, contradicting $A\subseteq U$. Thus the closure of a connected set is connected. The empty-set case holds directly.

Treat $0$ as the trivial group if $G$ is not abelian. The restriction maps satisfy the presheaf identities. To verify the sheaf gluing axiom, take an open cover $(V_i)$ of $U$. If $x\notin U$, every section group is trivial. If $x\in U$, at least one $V_i$ contains $x$. Every two such opens have an overlap containing $x$, so compatibility forces all their sections to have the same value $g\in G$. Opens not containing $x$ have their unique trivial section. The value $g\in \mathcal{G}(U)$ is exactly the unique glued section. Thus the stated presheaf is a skyscraper sheaf, including for nonabelian groups.

For the stalk of a sheaf, take the direct limit over neighbourhoods of $y$. If some neighbourhood $V$ of $y$ omits $x$, neighbourhoods inside $V$ are cofinal and all their groups are trivial. If every neighbourhood of $y$ contains $x$, every group in the system is $G$ and every transition map is the identity. Consequently, on the arbitrary topological space in the question,Membership in this closure means exactly that every neighbourhood of $y$ contains $x$. In particular ${\mathcal{G}}_x=G$. If $x$ is a closed point, the closure is just $\{x\}$ and the familiar point-supported answer results. The separation assumption is not present in the printed question, so it cannot be imposed silently.

Use the product topology on $\mathcal{C}=[k{]}^{\mathbb{Z}}$, with $[k]$ discrete, and the left shift $(\mathcal{L}c)(j)=c(j+1)$. A basic cylinder set specifies finitely many coordinates. This is a compact metric space, and $\mathcal{L}$ is a homeomorphism. Write $Y=\overline{\{{\mathcal{L}}^{n}c:n\ge 0\}}$ for the forward orbit closure. A minimal point means that $Y$ is a minimal dynamical system, equivalently that every point of $Y$ has a dense forward orbit in $Y$; it need not be a fixed point.

There is a convention needed in the source: the bounded-gaps property concerns every finite integer interval $I$, hence every finite word over an alphabet. Literally allowing $I=\mathbb{Z}$ would make the displayed condition impossible for finite $U$, although a constant coloring is a minimal point. Under the standard finite-word interpretation, the property is precisely uniform recurrence.

Suppose first that $Y$ is a minimal dynamical system. Its nonempty compact forward-invariant subset $\mathcal{L}(Y)$ must equal $Y$. As the ambient left shift is invertible, all integer translates of $c$ belong to $Y$. Let $I=[a,b]\cap \mathbb{Z}$ have length $\ell =b-a+1$, and let $V=\{z\in Y:z{|}_I=c{|}_I\}$. This is a nonempty clopen set. Each forward orbit in $Y$ meets $V$, so $\{{\mathcal{L}}^{-n}V:n\ge 0\}$ covers $Y$. By compactness, finitely many suffice; let $D$ be the largest index in this finite cover. For any integer $u$, the point ${\mathcal{L}}^{u-a}c$ lies in $Y$, so some $0\le n\le D$ satisfies ${\mathcal{L}}^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell -1]$, entirely inside $[u,u+D+\ell -1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.

Conversely suppose $c$ is uniformly recurrent. Every finite word from any integer translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete alphabet forces eventual exact agreement on that block. Given $I=[a,b]\cap \mathbb{Z}$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\ge 0$, so ${\mathcal{L}}^{n}z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap \mathbb{Z}$ for arbitrarily large $r$ proves that $c$ belongs to the forward orbit closure of every $z\in Y$. That closure is a closed forward-invariant subset of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward orbit is therefore dense in $Y$, provingThe same conclusion holds if orbit closure is defined using all integer iterates: the bounded-gaps condition makes the forward and two-sided closures equal.

Write $\mathbb{H}=\{z:\mathrm{Im}z>0\}$. A compact H-hull is a bounded, relatively closed set $A\subseteq \mathbb{H}$ for which $D_A=\mathbb{H}\setminus A$ is a simply connected domain. “Compact” refers to its compact closure in $\overline{\mathbb{H}}$; boundary points on $\mathbb{R}$ can be adjoined without changing the remaining domain.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:D_A\to \mathbb{H}$ with hydrodynamic normalization at infinity. By the Schwarz reflection principle outside a large disc, it has the Laurent seriesThe half-plane capacity is the coefficientThe empty compact H-hull has $g_A(z)=z$ and half-plane capacity zero. Nonnegativity follows from the Brownian representation of half-plane capacity proved next.

For $r>0$, the conformal map $h(z)=r{g}_A(z/r)$ maps $\mathbb{H}\setminus rA$ onto $\mathbb{H}$ and has Laurent series $h(z)=z+r^{2}a/z+O(z^{-2})$. The uniqueness in hydrodynamic normalization at infinity givesAt $r=0$ the scaled closure is contained in $\mathbb{R}$, so the interior compact H-hull is empty and both sides are zero. This is the scaling part of scaling and translation of half-plane capacity.

For the vertical slit, choose the branch of $\sqrt{z^2+1}$ asymptotic to $z$ at infinity. It is the mapping-out function of a compact H-hull for $(0,i]$, andThus half-plane capacity of a vertical slit givesThe endpoint $0$ is merely part of the boundary closure convention.

For the filled upper half-disc $K=\{z\in \mathbb{H}:|z|\le 1\}$, the conformal map $g_K(z)=z+1/z$ maps the exterior half-disc onto $\mathbb{H}$. Indeed, it sends the semicircle to $[-2,2]$, the remaining real boundary to the complementary intervals, and its inverse is the branch of $(w+\sqrt{w^2-4})/2$ asymptotic to $w$. Hence half-plane capacity of a half-disc givesIf $\mathbb{D}$ denotes the open unit disc, the printed $\mathbb{H}\cap \mathbb{D}$ is not relatively closed and is literally not a compact H-hull. The intended value is the one for its filled relative closure $K$. With a closed-disc convention the displayed notation already represents that hull.

For a nonempty compact H-hull $A$, its closure meets $\mathbb{R}$: otherwise a nonempty compact set strictly inside $\mathbb{H}$ would be separated from the lower half-plane in the complement, contradicting the simply connected domain condition for $\mathbb{H}\setminus A$. Choose $x\in \overline{A}\cap \mathbb{R}$ and put $d=\mathrm{diam}(A)$. Taking limits in the definition of diameter shows $|z-x|\le d$ for every $z\in A$. Thus $A$ lies in the filled half-disc of radius $d$ centered at $x$. Its half-plane capacity is $d^2$ by scaling and translation of half-plane capacity. Using monotonicity of half-plane capacity,For the empty compact H-hull, use diameter zero. This proves the unit-constant version of half-plane capacity is bounded by squared diameter; no claim of optimality of the constant is needed.