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In a plane domain, a closed discrete subset is exactly a set meeting every compact subset in finitely many points. This local finiteness is necessary for a nonzero holomorphic function to have the subset as its entire zero set. A merely isolated-point subset can have an accumulation point outside the subset but inside the domain, as $\{1/n:n\ge 1\}\subset \mathbb{C}$ does.

// Destination: analysis.bigb

Enumerate all intervals with rational endpoints, using the countability of the rational numbers and listing pairs of enumeration indices by increasing sum. If every point $b$ of $B$ is an isolated point, some such interval contains $b$ and no other point of $B$. Assign to $b$ the least index of such an interval. Distinct points cannot have the same index, so this gives an injective function into $\mathbb{N}$ and proves that $B$ is a countable set. The argument uses isolation in the subspace topology, not closedness: the discrete subset $\{1/n:n\ge 1\}$ illustrates that an accumulation point outside $B$ is allowed.

The zero-prescription claim requires a closed discrete subset, meaning no accumulation in $\Omega$. If “discrete” means only that each listed point is isolated, it is false: $\{1/n:n\ge 1\}\subset \mathbb{C}$ is a discrete subset, but any holomorphic function vanishing there is identically zero by the identity theorem for holomorphic functions, and then has additional zeros. In the usual zero-set interpretation, discreteness includes local finiteness. We prove the claim with this necessary interpretation.

First choose an increasing compact exhaustion $K_j$ of $\Omega$ by relatively filled sets: every component of $\hat{\mathbb{C}}\setminus K_j$ meets $\hat{\mathbb{C}}\setminus \Omega$. Such an exhaustion is obtained by exhausting with finite unions of small closed squares lying in $\Omega$, then adding all complementary components relatively compact in $\Omega$. The filled sets remain compact in $\Omega$: a complementary component containing a point within distance $\mathrm{dist}(K_j,\mathbb{C}\setminus \Omega )$ of the complement could join that complement through a ball disjoint from $K_j$, and hence would not be filled. Their boundedness follows by using a disk containing the initial compact. Nesting can be ensured by enlarging each stage first. This is also the geometric description of the holomorphic convex hull proved below.

A finite prescribed set is realized by a finite product of linear factors. For an infinite closed discrete subset, enumerate the distinct points as $a_n$. Every compact contains only finitely many of them, so there are indices $j(n)\to \infty$ with $a_n\notin K_{j(n)}$. Finitely many initial factors can be handled without an approximation requirement. For the remaining factors, let $U_n$ be the complementary component containing $a_n$. It also contains a point $c_n\notin \Omega$ or infinity. Join $a_n$ to $c_n$ by an arc in that component, or to infinity by a ray-like arc. DefineEach $p_n$ is a holomorphic function on $\Omega$, has exactly one simple zero, at $a_n$, and no other zeros there.

The arc is disjoint from $K_{j(n)}$, and $p_n$ has a holomorphic logarithm $L_n$ near that compact. To justify the logarithm, any closed curve avoiding the arc has equal winding numbers about its finite endpoints, so the integral of $p_n^{\prime }/p_n$ around it is zero. For a ray to infinity the winding number about $a_n$ is zero. A primitive of $p_n^{\prime }/p_n$, adjusted by a constant on each component, supplies the logarithm.

The Runge theorem proved in Question 4 approximates $L_n$ uniformly on $K_{j(n)}$ by $h_n\in \mathcal{O}(\Omega )$: all needed finite poles can be chosen outside $\Omega$ because every complementary component meets its complement. Choose the error small enough thatEach $F_n$ still has exactly the simple zero $a_n$. On every fixed compact, all sufficiently late factors satisfy the summable bound above. Hence their infinite product converges locally uniformly. It has a nonzero tail because, once $|F_n-1|<1/2$, its logarithmic tails satisfy $|\log F_n|\le 2|F_n-1|$ and converge absolutely. Thus multiplying by the finitely many earlier factors gives a holomorphic function with precisely the prescribed zeros, and no additional zeros. Repeating a factor to any prescribed finite positive multiplicity gives the corresponding multiplicity version. This is prescribed zeros in a plane domain via Runge approximation, not an invocation of the existence theorem being asked for.

For the specified infinite product, put $a_k=q^{-2k-1}$. On any compact annulus $0<r\le |z|\le R$,The logarithmic-tail estimate therefore proves locally uniform convergence on ${\mathbb{C}}^{\ast }$ and nonvanishing wherever no individual factor vanishes. At a zero of one factor, the product is zero and the remaining product converges to a nonzero value. ConsequentlyAll these zeros are simple and distinct: their moduli distinguish the two strings and their indices. At $z=0$ the reciprocal factors are undefined, so the given expression has no convergence value there. The infinite sequence of zeros tending to zero also rules out continuation across that point as a holomorphic function.

Set $P(z)={\prod }_{k\ge 1}(1-q^{-2k}z)(1-q^{-2k}{z}^{-1})$. Absolute local convergence allows reindexing:Since $1-z=-z(1-z^{-1})$, $\sigma (qz)=-z\sigma (q^{-1}z)$ for every $z\ne 0$. No division by a vanishing factor is involved, so the identity also holds at all the zeros.

A set is countable if it admits an injective function into $\mathbb{N}$, equivalently if it is finite or can be listed in a sequence. The empty set is included; an infinite countable set has a bijection with $\mathbb{N}$.

Every rational number has a representation $p/q$ with $p\in \mathbb{Z}$ and $q\ge 1$. For each $k\ge 1$, there are only finitely many such pairs with $|p|+q=k$. List these finite levels successively, ordering each level by $p$, and discard repeated fractions when they first recur. Every rational number appears in some finite level, proving $\boxed{\mathbb{Q}\text{ is countable}}$.

For the uncountability proof, suppose all numbers in $(0,1)$ could be listed as $r_1,r_2,\dots$. Choose each decimal expansion not to end in an infinite string of nines, and denote its $j$th digit by $d_{ij}$. Set $b_i=1$ when $d_{ii}\ne 1$, and $b_i=2$ otherwise. Then the decimaldefines a number in $(0,1)$ with a unique such expansion, since its digits are only $1$ and $2$. It differs from $r_i$ at digit $i$ for every $i$, contradicting the enumeration. This Cantor diagonal argument proves $\boxed{\mathbb{R}\text{ is uncountable}}$ because its subset $(0,1)$ already is.

For the union of two countable sets $C,D$, take injective functions ${\iota }_C:C\to \mathbb{N}$ and ${\iota }_D:D\to \mathbb{N}$. The mapis injective: it is injective within each branch, and their images have different parity. Thus $\boxed{C\cup D\text{ is countable}}$, regardless of overlap.

The property of $A$ says that both $A$ and its complement are dense subsets of the real line. Both possible cardinalities occur. To establish the needed density directly, given $x\in \mathbb{R}$ and $\epsilon >0$, choose an integer $q>1/\epsilon$. Then $\lfloor qx\rfloor /q$ is rational and differs from $x$ by less than $1/q<\epsilon$. For an irrational approximation, choose a rational $r$ within $\epsilon /2$ of $x$ and a positive integer $q$ large enough that $\sqrt{2}/q<\epsilon /2$. The number $r+\sqrt{2}/q$ is irrational and lies within $\epsilon$ of $x$; irrationality follows from that of $\sqrt{2}$, since otherwise subtracting $r$ and multiplying by $q$ would make $\sqrt{2}$ rational. To recall the latter fact, a reduced fraction $p/q=\sqrt{2}$ would satisfy $p^2=2q^2$, forcing $p$ even and then $q$ even, contrary to coprimality. ThereforeThe second set is uncountable because, if it were countable, its union with $\mathbb{Q}$ would make $\mathbb{R}$ countable. The same two density arguments give the required approximations from $A$ and from its complement in either example.

Finally, the condition on $B$ makes every one of its points an isolated point, so $B$ is a discrete subset of the real line. Enumerate all rational intervals $(p_j,q_j)$ with $p_j<q_j$; enumerate pairs of indices of the rational list by increasing sum and keep those whose endpoints have the required order. For each $b\in B$, its isolating neighbourhood contains a rational interval $(p_j,q_j)$ containing $b$ and no other point of $B$. Assign to $b$ the least such index $j$. Two distinct points cannot receive the same index, because that interval would then contain two members of $B$. This gives an injective function $B\to \mathbb{N}$ and proves that discrete subsets of the real line are countable:Using the least index makes the assignment explicit and requires no arbitrary selection of an uncountable family of intervals.