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If $\mathcal{B}$ is a countable topological basis of $X$, then $\{B\cap Y:B\in \mathcal{B}\}$ is a countable topological basis of the subspace $Y$. Indeed every subspace-open neighborhood is $Y\cap U$ for an open $U$ of $X$, and a suitable basic set lies between its point and $U$. This proves the first assertion.

Urysohn's metrisation theorem: every second-countable space that is also a T3 space is a metrizable space. Here regularity of a topological space allows a neighborhood of a point to be shrunk to an open neighborhood whose closure is still inside the original one.

We first prove the separation facts needed in a complete proof. Second countability implies the Lindelöf property: for each basic set contained in some member of an open cover choose one such member; these countably many members cover. In particular closed subspaces are Lindelöf. For disjoint closed sets $F,G$, regularity of a topological space and this property give open covers $F\subseteq {\bigcup }_n{U}_n$, $G\subseteq {\bigcup }_n{V}_n$ with ${\overline{U}}_n\cap G=\varnothing$ and ${\overline{V}}_n\cap F=\varnothing$. DefineThese sets are open and contain $F,G$ respectively. They are disjoint: if a point belonged to a term indexed by $n$ in the first union and by $m$ in the second, the term with the larger index would have removed the closure of the other covering set. Thus the space is a normal space.

For completeness, Urysohn's lemma supplies continuous separating functions in a normal topological space. Its construction is as follows. Given disjoint closed $F,G$, choose open sets $O_r$ for dyadic $r\in [0,1]$, with $F\subseteq O_0$, $O_1=X\setminus G$, and ${\overline{O}}_r\subseteq O_s$ whenever $r<s$. Start by shrinking a neighborhood of $F$ inside $X\setminus G$, and insert the successive dyadic sets by the closed-neighborhood shrinking in a normal space. DefineThen $f=0$ on $F$, $f=1$ on $G$, and $0\le f\le 1$. The identities $\{f<a\}={\bigcup }_{r<a}{O}_r$ and $\{f>a\}={\bigcup }_{a<r<1}(X\setminus {\overline{O}}_r)$ prove continuity for $0<a<1$, with the endpoint cases immediate.

Let $\{B_n\}$ be a countable topological basis. For each pair $(n,m)$ with ${\overline{B}}_n\subseteq B_m$, choose a continuous function $f_{nm}:X\to [0,1]$ equal to one on ${\overline{B}}_n$ and zero outside $B_m$. These form a countable family. Given $x\in O$ open, choose $B_m$ with $x\in B_m\subseteq O$, shrink a neighborhood of $x$ inside $B_m$, and then choose $B_n$ inside that smaller neighborhood. ThusConsequently these functions generate the original topology. They also separate distinct points: use the $T_1$ open set $X\setminus \{y\}$ when $x\ne y$. Enumerate them as $f_j$, padding by zero functions if necessary. The map $x\mapsto (f_j(x){)}_j$ is a topological embedding into the countable cube. Pulling back its metric, or directly usinggives a compatible metric. The fact that the functions distinguish distinct points proves positivity for distinct points, the triangle inequality is termwise, and the uniformly bounded tails show that this metric induces exactly the topology generated by the coordinates. Empty and singleton spaces are trivially metrizable spaces. This proves the theorem.

For the application, a separable metric space has a countable topological basis consisting of balls with rational positive radii about a countable dense set, and it is a T3 space. The given product theorems make a countable product of these spaces second countable and a T3 space. Urysohn's theorem makes it a metrizable space. Finally every second countable space is separable: choose a point from each nonempty basic open set; the resulting countable set meets every nonempty open set. Hence the product is separable and a metrizable space.

Write $\alpha X=X\cup \{{\infty }_X\}$. Its open sets are the open subsets of $X$ together with sets $\alpha X\setminus K$, where $K$ is a compact subset of $X$. Compact subsets of the Hausdorff space $X$ are closed. This fact, and finite unions of compact sets, verify the topology axioms, including intersections of an old open set with a neighborhood of infinity. The inclusion $h_1:X\to \alpha X$ is a topological embedding with open image.

An open cover of $\alpha X$ has a member containing infinity; its complement is compact in $X$ and therefore is covered by finitely many other members. This proves compactness. Two points of $X$ can be separated in $X$. A point $x$ and infinity can be separated by an open neighborhood $U$ of $x$ contained in a compact neighborhood $K$, and $\alpha X\setminus K$. Thus $\alpha X$ is Hausdorff. Since $X$ is noncompact, every neighborhood of infinity meets $X$, so the embedded copy of $X$ is dense. This proves it is a Hausdorff compactification.

For a proper map, use the standard locally compact Hausdorff convention: it is continuous and inverse images of compact sets are compact. DefineAn open subset of $Y$ has open inverse image in $X$, hence in $\alpha X$. A basic neighborhood $\alpha Y\setminus K$ of infinity has inverse image $\alpha X\setminus f^{-1}(K)$, open because $f^{-1}(K)$ is compact. Hence $\alpha f$ is continuous and the diagram commutes. Any continuous map from a compact space to a Hausdorff space is proper: a compact subset of the target is closed, its inverse image is closed in the compact domain and therefore compact. This proves the required properness.

The last requested extension does not exist, even as a continuous map. Both sequences $h_1(0,n)$ and $h_1(1,n)$ tend to ${\infty }_{{\mathbb{R}}^2}$, since they eventually leave every compact subset of the plane. Their prescribed images would be the constant sequences $h_2(0)$ and $h_2(1)$. Continuity would force both distinct points to equal $g({\infty }_{{\mathbb{R}}^2})$, contradicting uniqueness of limits in the Hausdorff compactification of the line. The original product projection is not proper, as its inverse image of a singleton is an unbounded line.

For $x$ in a compact Hausdorff space $K$, its quasi-component $Q$ is the intersection of all clopen sets containing $x$. It is closed and therefore compact. Suppose $Q=A\bigsqcup B$ is a separation of a topological space into two nonempty relatively closed sets, with $x\in A$. The sets $A,B$ are compact and hence closed in $K$. Normality of a compact Hausdorff space gives disjoint open neighborhoods $U,V$ of $A,B$.

The compact set $K\setminus (U\cup V)$ is disjoint from $Q$. For each point $y$ of that set choose a clopen $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus (U\cup V)$, so their corresponding intersection $C$ is clopen, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is clopen in $C$, hence clopen in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus $Q$ is connected. Any connected set containing $x$ lies in every clopen neighborhood of $x$, so the connected component of $x$ equals $Q$.

For the second assertion, $D$ is closed in the ambient Hausdorff space $X$, and is itself compact Hausdorff. Let $C$ be a connected component of $D$ disjoint from ${\partial }_{X}D$. Choose $p\in C$. By the first assertion $C$ is the quasi-component of $p$ in $D$. For each $y\in {\partial }_{X}D$ choose a clopen subset of $D$ containing $p$ but not $y$. Compactness of ${\partial }_{X}D\subseteq D$ gives a finite intersection $H$ of these sets which contains $C$ and avoids the whole boundary. Thus $H\subseteq {\mathrm{int}}_{X}D$. It is closed in $X$, because it is closed in the closed set $D$, and is open in $X$, because it is relatively open in $D$ and contained in its interior. It is nonempty and proper, contradicting connectedness of $X$. Hence every connected component of $D$ meets its boundary.

Finally suppose a closed subset $A$ of the square has the point-separating clopen property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial line segment: it contains an interior point of the square and a small ball about that point. If such a segment lay in $A$, a clopen subset of $A$ separating its endpoints would induce a separation of a topological space of the connected segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is nowhere dense in the square, and also nowhere dense in ${\mathbb{R}}^2$.

Use these standard ultrafilter facts: a space is compact if and only if every ultrafilter on it converges; the image ultrafilter under a map is an ultrafilter; and a convergent ultrafilter contains every neighborhood of its limit.

Let $X={\prod }_{\lambda \in \Lambda }{X}_{\lambda }$ with all factors compact. An empty factor makes $X$ empty and hence compact. Otherwise, given an ultrafilter $\mathcal{U}$ on $X$, push it forward by each product projection to obtain an ultrafilter ${\mathcal{U}}_{\lambda }$ on $X_{\lambda }$. Choose a limit $x_{\lambda }$ of each, and put $x=(x_{\lambda }{)}_{\lambda }$. A basic product neighborhood of $x$ is an intersection of finitely many inverse images of coordinate neighborhoods. Each inverse image belongs to $\mathcal{U}$, so their finite intersection does too. Thus $\mathcal{U}$ converges to $x$, proving compactness of the product. Uniqueness of coordinate limits is not needed, so the factors need not be Hausdorff.

For local compactness, suppose only the factors indexed by a finite set $F$ may be noncompact. At $x\in X$ choose, for each $\lambda \in F$, a compact neighborhood $K_{\lambda }$ of $x_{\lambda }$, with an open neighborhood $U_{\lambda }\subseteq K_{\lambda }$. For other indices put $K_{\lambda }=X_{\lambda }$. The compact product ${\prod }_{\lambda }{K}_{\lambda }$ contains the open product neighborhood restricting just the finitely many coordinates in $F$ to $U_{\lambda }$. Hence every point has a compact neighborhood and the product is locally compact.

The unrestricted assertion is false. Each factor of ${\mathbb{R}}^{\mathbb{N}}$ is locally compact, but its product is not. Any basic neighborhood restricts only finitely many coordinates. Choose a free coordinate $j$; its projection of that neighborhood is all of $\mathbb{R}$. A compact neighborhood containing it would consequently have compact image equal to $\mathbb{R}$, impossible. This shows directly that no point has a compact neighborhood.

Complete the square-class map at the identity by $q(O)=1$; at the finite point $T=(0,0)$ its value is the class of $b$. For another rational point writeThese denominator powers follow because a negative valuation of $x$ forces the leading cubic term to dominate, so that valuation is even and the valuation of $y$ is three halves of it. The equation givesIf a prime $\ell \nmid b$ divides $r$, it does not divide the second factor, since $\ell \nmid s$. Thus $v_{\ell }(r)$ is even. All denominator valuations are even too. Every image class therefore has a representative which is a signed squarefree divisor of $b$, proving that $\mathrm{im}q$ is finite.

Use the standard 2-isogeny with kernel $\{O,T\}$ and dual isogeny satisfying $\hat{\varphi }\varphi =[2]$, $\varphi \hat{\varphi }=[2]$. Away from the exceptional points one may takeApplying the analogous isogeny to $E^{\prime }$ gives a curve with coefficients $4a,16b$; scaling its coordinates by $1/4,1/8$ gives the dual map back to $E$. The analogous descent map $q^{\prime }$ on $D$ has kernel $\varphi (C)$ and finite image. The sequenceis exact at the last two terms. Its outside groups are finite, identified with the two finite descent images, so $C/2C$ is finite.

For $E_p:y^2=x(x^2-p^2)$, the rational 2-torsion is $O,(0,0),(p,0),(-p,0)$, of order four. The first descent image is exactly $\{1,-1,p,-p\}$: these classes occur on the 2-torsion, and the divisor-support bound permits no others. The isogenous curve is $E_p^{\prime }:y^2=x(x^2+4p^2)$. Its real points have $x\ge 0$, so the second image lies in the four positive classes $\{1,2,p,2p\}$.

Here the kernel $\{O,(0,0)\}$ of the dual is contained in $\varphi (C)$, since $\varphi (\pm p,0)=(0,0)$. Thus the displayed map $D/\varphi (C)\to C/2C$ is injective, and exactness givesThe allowed rank formula makes the left side $2^{R+2}$, soNow consider the quartic relation. Normalize a nontrivial integer solution so that $p$ does not divide both $x,y$, by dividing them by $p$ and $z$ by $p^2$ whenever necessary. If $p\nmid x$, reduction modulo $p$ gives $x^4=2z^2$, with $z\ne 0$ modulo $p$, making $2$ a quadratic residue. If $p\mid x$, then $p\nmid y$ and $p\mid z$; write $x=pX$, $z=pZ$, divide by $p^2$, and reduce to get $y^4=2Z^2$ modulo $p$. Again $2$ is a square. ThereforeIf the class $2$ occurred in $\mathrm{im}{q}^{\prime }$, write the corresponding coordinate as $x^{\prime }=2u^2/v^2$. The curve equation then givesHere $u,v$ are nonzero coprime integers and $z$ is rational. Since $2z^2$ is an integer, its reduced denominator must be one, so $z$ is an integer too. For $p\equiv 3,5(\mathrm{mod}8)$ this contradicts the quartic obstruction. Hence class $2$ is absent. A subgroup of $\{1,2,p,2p\}$ omitting $2$ has order at most two, so $2^R\le 2$ and $R\le 1$.

For $p=5$, direct substitution verifies the nonintegral pointA rational torsion point would have integral coordinates by the preceding theorem, so this point has infinite order. Thus $R\ge 1$, and together with the upper bound,

The vertical integral is absolutely convergent because the denominator is quadratic in the imaginary part. If $x<1$, move its line to the right: no poles are crossed, and the integral on $\mathrm{Re}s=R$ is $O(x^R/R)$, tending to zero. If $x>1$, move left across the simple poles at $0,-1$. Their residues are $1$ and $-x^{-1}$, and the far-left integral tends to zero. Horizontal segments vanish by their $O(|\mathrm{Im}s{|}^{-2})$ decay for a fixed pair of vertical lines. At $x=1$, dominated convergence on the original line gives the limiting value zero. Hence the kernel is $0$ below one and $1-x^{-1}$ at or above one.

On a line $c>2$, absolute convergence permits interchanging the totient Dirichlet series and this integral. The $n$th term is $x\varphi (n)$ times the kernel at $x/n$, giving exactly $\varphi (n)(x-n)$ for $n\le x$. ThusWe may take $c=5/2$ and shift to $\mathrm{Re}s=3/2$. The denominator has no zeros in this strip; the only pole crossed is $s=2$, where $\zeta (s-1)$ has residue one. Its residue in the integrand isOn the new line, the supplied estimates bound the integrand by $O(x^{5/2}|t{|}^{-3/2}{\log }^8|t|)$ for $|t|\ge 2$, an integrable function. The compact portion is bounded as well. The same estimates, uniformly across the strip, make the horizontal integrals tend to zero. ConsequentlyFor the Tauberian step let $A(x)={\sum }_{n\le x}\varphi (n)$. It is nondecreasing because the summands are nonnegative, and $S(x)={\int }_0^{x}A(t)dt$. For fixed $0<ϵ<1$,After division by $x^2$ and passage to lower and upper limits, the asymptotic for $S$ gives bounds $(3-3ϵ+{ϵ}^2)/{\pi }^2$ and $(3+3ϵ+{ϵ}^2)/{\pi }^2$. Letting $ϵ\downarrow 0$ squeezes them together. ThereforeThis uses monotonicity of the summatory function; differentiating an asymptotic formula without such control would not justify the conclusion.

Use the convention $f(1)=1$ in the definition of a multiplicative arithmetic function. For a finite set of primes, multiplying the absolutely convergent local series gives the Dirichlet sum over integers supported on those primes, by unique prime factorization. If all primes up to $Y$ are included, any missing integer is larger than $Y$, so the difference from the full series is bounded by ${\sum }_{n>Y}|f(n)|n^{-\sigma }$, which tends to zero. This proves the Euler product in its absolute-convergence half-plane.

For coprime $m,n$, every divisor of $mn$ is uniquely a product of a divisor of $m$ and a divisor of $n$. Hence ${\sigma }_{\alpha }(mn)={\sigma }_{\alpha }(m){\sigma }_{\alpha }(n)$. The squared modulus is also multiplicative when $\alpha =i\theta$.

At a prime put $z=p^{-s}$ and $a=p^{i\theta }$, so $|a|=1$. Summing by the two divisor exponents givesFor example, expand the square as ${\sum }_{j,l\le k}{a}^{j-l}$ and sum $z^k$ from $k=\max (j,l)$ to obtain this rational expression. It remains valid for $a=1$. Absolute convergence for $\sigma >1$ follows from $|{\sigma }_{i\theta }(p^k)|\le k+1$ and convergence of ${\sum }_p{p}^{-\sigma }$. Multiplying the factors therefore provesSuppose now that $\zeta (1+i\theta )=0$. Necessarily $\theta \ne 0$, since $s=1$ is a pole. Complex conjugation gives a zero also at $1-i\theta$. In the numerator, the double pole at $s=1$ is canceled by those two zeros in the shifted factors. At $s=1\pm i\theta$, the possible simple pole of a shifted factor is canceled by the double zero of $\zeta (s{)}^2$. Thus on $\mathrm{Re}s>0$ the only possible remaining poles come from zeros of $\zeta (2s)$. Since the Euler product excludes zeros for $\mathrm{Re}(2s)>1$, these possible poles lie in $0<\mathrm{Re}s\le 1/2$. In particular $f$ is regular for $\mathrm{Re}s>1/2$.

The assumed nonnegative-series principle now makes its Dirichlet series converge and equal $f$ there. Its first coefficient is one and all coefficients are nonnegative, so $f(\sigma )\ge 1$ for real $\sigma >1/2$. But as $\sigma \downarrow 1/2$, the numerator remains finite while $\zeta (2\sigma )\to +\infty$, forcing $f(\sigma )\to 0$. This contradiction provesAt the real point $s=1$ the statement is understood as a pole rather than a finite nonzero value. The pole-location argument did not assume the zero-free boundary before proving it.

For $N=\lfloor x\rfloor$, interchange a finite sum and an integral:This proves the Abel's summation formula, including the term at $n=1$.

Apply it with $f(n)=1$, $F(t)=1/t$. Writing $\{t\}=t-\lfloor t\rfloor$ givesThe last two terms are $O(x^{-1})$, proving the first expansion. For the refinement let $B_1(t)=\{t\}-1/2$. It has mean zero on each unit interval, so its primitive $P(t)={\int }_0^t{B}_1(u)du$ is bounded and periodic. Integration by parts givesThus for integer $x$, where $\{x\}=0$,For the zeta representation, integrating each defining integral by parts yieldsSumming from $1$ to $N-1$ and substituting into the definition of $\zeta$ givesSince $0\le x-n\le 1$,This is uniform on compact subsets of $\mathrm{Re}s>0$, so the series is holomorphic there and limits at $s=1$ are legitimate after subtracting the pole. Expanding $N^{1-s}=1-(s-1)\log N+O((s-1{)}^2)$ givesThe final tail is $O(N^{-1})$, and the harmonic expansion sends the remaining difference to $\gamma$. Therefore