The empty set is bounded. Otherwise choose $p$ in the compact metric space $X$. The open balls $B(p,n)$, $n=1,2,\dots$, form an open cover. A finite subcover exists by compactness; because the balls are nested, the largest one already covers $X$. Hence $X\subseteq B(p,N)$ for some $N$.

For $x,y\in X$, the triangle inequality gives $d(x,y)\le d(x,p)+d(p,y)<2N$. ThusThe proof uses only compactness and the metric, not a coordinate system.

The Heine-Borel theorem says that a subset $K$ of finite-dimensional Euclidean space is compact exactly when it is both closed and bounded:Here compactness refers to the subspace topology induced by the usual Euclidean metric, and closedness and boundedness are measured in the ambient Euclidean space. This characterization does not hold for arbitrary metric spaces.

Consider the counterclockwise unit circle and setThe parametrization $z=e^{i\theta }$ givesThe imaginary part is zero: replacing $\theta$ by $2\pi -\theta$ conjugates the integrand, since $n$ is an integer and $x$ is real. The real part is precisely the defining Bessel integral, so $I=J_n(x)$.

Both exponential series converge absolutely and uniformly on the circle. Their product may therefore be integrated term by term:Uniform absolute convergence can also be checked by summing the absolute coefficient bounds, which gives $e^{|x|}$. The residue is obtained exactly when $n-1+j-k=-1$, or $k=n+j$. The Cauchy coefficient formula consequently yieldsThis proves the Bessel function of the first kind series directly from its integral, including the sign and factorial normalization.

A simple zero means $h(a)=0$ and $h^{\prime }(a)\ne 0$. The local factorization is $h(z)=(z-a)h_1(z)$, where $h_1$ is holomorphic and $h_1(a)=h^{\prime }(a)\ne 0$. ThusThe numerator is holomorphic near $a$, so its constant Taylor coefficient is the coefficient of $(z-a{)}^{-1}$ in the quotient. ThereforeIf $g(a)=0$, the possible pole cancels and the residue is zero; the formula does not require $g/h$ actually to have a pole. This is the residue formula for a quotient at a simple denominator zero.

Write the Laurent series about $a$ as ${\sum }_{k\in \mathbb{Z}}{c}_k(z-a{)}^k$. The isolated singularity is:

These cases exhaust the possible principal parts. The residue isFor the first example, $\sin z=z+O(z^3)$ givesThus zero is a double pole, with residue zero. For the other example,so the ratio is $-z/2+O(z^2)$. Its singularity is removable, with extended value zero and residue zero. The numerator's zero has higher order than the denominator's, so there is no pole.

For fixed $z\notin \mathbb{Z}$, integratecounterclockwise around the specified square of half-side $R_N=N+1/2$, choosing $N>|z|$. The integer poles inside it are $-N,\dots ,N$, and their residues are $1/(n^2-z^2)$ because $\pi \cot (\pi w)$ has residue one at each integer.

The two other possible poles are $w=z$ and $w=-z$. Their residues are each $\pi \cot (\pi z)/(2z)$, hence sum to $\pi \cot (\pi z)/z$. This formula still applies when the numerator vanishes and a possible pole is removable. The residue theorem givesOn the square, $|w|\ge R_N$, its perimeter is $8R_N$, and the supplied cotangent bound givesThe series converges absolutely by part (b), so its symmetric partial sums have the same limit as the full sum. ThereforeAt half-integers the cotangent form makes the value zero explicit; the reciprocal-tangent notation is understood by its continuous extension there. The cotangent contour summation rests on the contour estimate, not only on formally matching poles.

Fix any bounded disk avoiding the integers, and choose $R$ with $|z|\le R$ on it. For sufficiently large $|n|$, say $|n|\ge 2R$ and $|n|\ge 1$,The convergent numerical series ${\sum }_{n\ne 0}{n}^{-2}$ and the Weierstrass M-test give absolute and uniform convergence of the tails. The finitely many remaining summands are holomorphic away from their integer poles. Thus the partial sums converge locally uniformly on $\mathbb{C}\setminus \mathbb{Z}$.

By part (a)(iii), the sum is a holomorphic function on $\mathbb{C}\setminus \mathbb{Z}$. The term $n=0$ is $-1/z^2$ and is harmless on this domain; it must not be omitted.