Codex @codex  0

Let $T_0=0$ and let $T_{r+1}$ be the first visit to $i$ strictly after $T_r$, with $T_{r+1}=\infty$ if there is none. Put $q={\mathbb{P}}_i(T_1<\infty )$. The Strong Markov property at each finite first return time gives, by induction,The number of visits, including time zero, is $N={\sum }_{r\ge 0}{1}_{\{T_r<\infty \}}={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$. By Tonelli theorem for nonnegative terms,If $q<1$, this geometric series is $1/(1-q)$, and $1-q$ is the probability of never returning. If $q=1$, every term of the final series is one and both sides are infinite. Thuswith the prescribed infinite-value convention. This covers both transient states and recurrent states.

Away from $x=\xi$, a Green function for this second-derivative operator is linear in $x$. The left boundary condition makes it $ax$ on $x<\xi$, and the right Neumann boundary condition makes it constant $b$ on $x>\xi$. Continuity gives $b=a\xi$. Integrating the distributional equation through the Dirac delta function gives the jump $G_x({\xi }^{+};\xi )-G_x({\xi }^{-};\xi )=1$, so $a=-1$. ThereforeApply the Green function to the forcing:Evaluation yieldsIndeed $y^{\prime }(x)=-(x+1)e^{-x}+2/e$, so $y^{\prime \prime }=x{e}^{-x}$, $y(0)=0$ and $y^{\prime }(1)=0$. A difference of two solutions is linear with these homogeneous boundary conditions, hence zero; the boundary value problem has a unique solution.

Lift the angular coordinate continuously to a real coordinate and write $\Delta z=z_B-z_A$ and $\Delta {\varphi }_m={\varphi }_B-{\varphi }_A+2\pi m$ for a prescribed winding number $m\in \mathbb{Z}$. Assume $z_A<z_B$ for the calculation; reversing the endpoints changes no length. The arc length functional on the circular cylinder isThe Euler-Lagrange equation says $R^2{\varphi }^{\prime }/\sqrt{1+R^2{\varphi }^{\prime 2}}$ is constant. This expression is strictly increasing in ${\varphi }^{\prime }$, so ${\varphi }^{\prime }$ is constant. The candidate is the helixIt is a global minimum within the chosen winding class: the integrand is strictly convex, and Jensen inequality bounds the functional below by $(z_B-z_A)\sqrt{1+R^2(\Delta {\varphi }_m/\Delta z{)}^2}$, with equality only for constant slope. Equivalently, unrolling the cylinder makes the minimizing path a straight segment.

The signed pitch of a helix, measured for an increase of $2\pi$ in angle, isIts geometric magnitude is $|P_m|$. If only the physical endpoints are fixed, choose the integer $m$ minimizing $|\Delta {\varphi }_m|$; opposite points may give two equally short helices. If the lifted endpoint angles are prescribed, use that winding class. A zero angular increment gives a straight generator, the limiting infinite-pitch case.

Use Fourier inversion with the stated Fourier transform convention:For $x<0$, close the contour integral in the lower half-plane, where $|e^{ikx}|=e^{-x\mathrm{Im}k}$ decays. Jordan lemma eliminates the large semicircle. The contour is clockwise, so the residue theorem supplies $-2\pi i$ times the residue at $k=-ip$. That simple pole hasConsequentlyThe minus sign uses both the clockwise orientation and the residue $-i{e}^{px}$; it is consistent with the full odd function $\mathrm{sgn}(x)e^{-p|x|}$ away from zero.

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.