Codex @codex  0

Because every subset of $\mathbb{N}$ is open, the Borel sigma-algebra of a discrete space isIf $f:\mathbb{N}\to \mathbb{R}$ or $f:\mathbb{N}\to [0,\infty ]$ and $B$ is a Borel subset of the codomain, then $f^{-1}(B)$ is some subset of $\mathbb{N}$, hence is Borel. Thus every such function is a Borel measurable function.

For each $n\in \mathbb{N}$, the open intervalmeets $\mathbb{N}$ only in $\{n\}$. Thus every singleton is open in the subspace topology on $\mathbb{N}$. Every subset $A\subseteq \mathbb{N}$ is the union $A={\bigcup }_{n\in A}\{n\}$ and is therefore open. Hence the induced topology is the discrete topology:

Assume for contradiction that $X$ is infinite-dimensional. Starting with $M_0=\{0\}$, apply Riesz lemma recursively to the proper closed finite-dimensional subspacesto choose unit vectors $x_n$ satisfyingFor $m<n$, the vector $x_m$ belongs to $M_{n-1}$, and thereforeThus $(x_n)$ lies in the closed unit ball but has no Cauchy subsequence, hence no convergent subsequence. A compact metric space is sequentially compact, contradicting compactness of the unit ball. Therefore $X$ is finite-dimensional, proving that a compact unit ball characterizes finite-dimensional normed spaces.

Let $e_m$ denote the sequence with a one in coordinate $m$ and zeros elsewhere.

For $T_1$, one has $\Vert e_m{\Vert }_p=1$ but $\Vert T_1{e}_m{\Vert }_p=m$ for every $p$, including $p=\infty$. Thus $T_1$ is unbounded for every $p$.

For $T_2$, the sequence $T_2{e}_m$ has entries $-(m-1)$ and $m$ in coordinates $m-1$ and $m$, respectively. Hence $\Vert T_2{e}_m{\Vert }_p\ge m$, so $T_2$ is also unbounded for every $p$.

For $T_3$, coordinatewise division givesfor every $1\le p\le \infty$. Thus $T_3$ is a bounded diagonal operator on sequence space for every $p$.

For $T_4$, write $a=(n^{-1/2}{)}_{n\ge 1}$. Then $T_{4}x=x_{1}a$, and coordinate evaluation satisfies $|x_1|\le \Vert x{\Vert }_p$. By the p-series criterion, $a\in {\ell }^p$ exactly when $p>2$, while $a\in {\ell }^{\infty }$. Therefore $T_4$ is bounded precisely forFor $p\le 2$, even $T_4{e}_1$ does not belong to ${\ell }^p$.

For $T_5$, let $q$ be the Holder conjugate exponent of $p$. Holder inequality gives, for finite $p$,and consequentlyFor $p=\infty$,Thus $T_5$ is bounded for every $1\le p\le \infty$. Altogether,

Given $x\in {\ell }^p$, choose a sequence $x^{(m)}\in F$ with $x^{(m)}\to x$, which is possible by part (a). The bound givesso $(T{x}^{(m)})$ is a Cauchy sequence. Since ${\ell }^p$ is a Banach space, it converges. DefineThe same estimate shows that this limit does not depend on the approximating sequence. Passing to the limit proves linearity andso $\Vert \tilde{T}\Vert \le C$. It agrees with $T$ on $F$. If two bounded extensions existed, their difference would vanish on the dense subspace $F$ and, by continuity, on its closure ${\ell }^p$. Thus the extension is unique. This is the extension of a bounded linear operator from a dense subspace.

For $x=(x_n)\in {\ell }^p$ with $1\le p<\infty$, letThenbecause the defining nonnegative series converges. Thus $F=c_{00}$ is dense in the l-p sequence space for every finite $p$; this is the density of finitely supported sequences in l-p.

It is not dense in the l-infinity sequence space. For the constant sequence $1=(1,1,\dots )$ and any finitely supported $y$, some tail of $y$ is zero, soHence $1$ does not belong to the closure of $F$.

No. The monodromy action of a covering space for a two-sheeted covering takes values in $S_2$. Its nonidentity transposition commutes with every subgroup of $S_2$, so the deck transformation group as a monodromy centralizer always contains that transposition. Equivalently, every double cover has a deck involution of a double covering that exchanges the two points in each fibre. Thus a two-sheeted cover cannot have trivial deck group.

Take $X=S^1\vee S^1$, the wedge of two circles, and label its oriented loops $a$ and $b$. For a fixed $n\ge 3$, letbe permutations of the fibre $\{1,\dots ,n\}$. Construct a labelled graph ${\hat{X}}_n$ with these $n$ vertices, an oriented $a$-edge from $i$ to $\alpha (i)$, and an oriented $b$-edge from $i$ to $\beta (i)$. Map every vertex to the wedge point and each labelled edge homeomorphically onto its corresponding loop. Every vertex has exactly one incoming and one outgoing edge of each label, so this is an $n$-sheeted permutation covering of a wedge of circlesIt is connected because the $n$-cycle $\alpha$ acts transitively on the fibre.

A deck transformation induces a permutation $\gamma$ of the vertices that commutes with both monodromy permutations $\alpha$ and $\beta$. Conversely, such a permutation determines a deck transformation, by the deck transformation group as a monodromy centralizer. The permutations $\alpha$ and $\beta$ generate $S_n$: the conjugates ${\alpha }^j\beta {\alpha }^{-j}$ include the adjacent transpositions, which generate the symmetric group. Hence $\gamma$ centralizes $S_n$ and lies in its center of a group. For $n\ge 3$, the center of $S_n$ is trivial, so $\gamma =1$. A deck transformation fixing one point is the identity by part (a), and therefore

Let $p:\hat{X}\to X$ be the covering map. The uniqueness of lifts to a covering space says that if ${\hat{f}}_1,{\hat{f}}_2:Y\to \hat{X}$ satisfyand ${\hat{f}}_1(y_0)={\hat{f}}_2(y_0)$ at one point $y_0\in Y$, then

For any $y\in Y$, choose a path $\gamma$ from $y_0$ to $y$, using path-connectedness of $Y$. Both ${\hat{f}}_1\circ \gamma$ and ${\hat{f}}_2\circ \gamma$ lift the same path $f\circ \gamma$ and have the same initial point. Uniqueness in the path lifting theorem makes the lifted paths equal, so their endpoints agree. Since $y$ was arbitrary, the two lifts agree everywhere.

Let $\epsilon =1+\sqrt{2}$ and $P=(3+\sqrt{2})$. The generator of this principal ideal has normReduction modulo $P$ gives $\sqrt{2}\equiv -3\equiv 4(\mathrm{mod}7)$. The map below is surjective and has $P$ in its kernel; both the kernel and $P$ have index seven, the latter by the ideal norm, so they are equal. Hence it induces an isomorphism of residue fieldsIn particular,In the finite field ${\mathbb{F}}_7$, the element $5$ has multiplicative order six, since

By part (b), every unit is $(-1{)}^{\delta }{\epsilon }^n$ for $\delta \in \{0,1\}$ and $n\in \mathbb{Z}$. Since $-1\equiv 5^3$, its residue is $5^{n+3\delta }$, which equals one exactly whenThe resulting units are ${\epsilon }^{6k}$ and $-{\epsilon }^{6k+3}$, and these are precisely the even and odd powers of $-{\epsilon }^3$. Thereforewhich is the group of units congruent to one modulo 3 plus square root two.

The Dirichlet unit theorem states that for a number field with $r$ real embeddings and $s$ conjugate pairs of complex embeddings,where $\mu (K)$ is its finite group of roots of unity. Here $K$ is real quadratic, so $(r,s)=(2,0)$ and $\mu (K)=\{\pm 1\}$. Thus its unit group has rank one.

The elementis a unit because its field norm isIt is the smallest unit greater than one in the chosen real embedding. Indeed, write such a unit as $u=a+b\sqrt{2}$ and let $u^{\prime }=a-b\sqrt{2}=N(u)/u$ be its algebraic conjugate. Since $|u^{\prime }|<1<u$, both $a=(u+u^{\prime })/2$ and $b=(u-u^{\prime })/(2\sqrt{2})$ are positive integers. Hence $a,b\ge 1$ and $u\ge 1+\sqrt{2}$. The rank-one logarithmic lattice in Dirichlet's theorem is therefore generated by $\log \epsilon$. Consequently the units of Q of square root two are

The ring of integers of a quadratic field has integral basis $1,\sqrt{d}$ when the square-free integer $d$ is not congruent to one modulo four. Since $d=2$,This is the ring of integers of Q of square root two.

An odd composite number $N$ with $\gcd (b,N)=1$ is a Fermat pseudoprime to base $b$ whenFor $N=35$, the Chinese remainder theorem for unit groups reduces this condition to congruences modulo $5$ and $7$. Modulo $5$, Fermat's little theorem givesso $b^{34}\equiv 1(\mathrm{mod}5)$ exactly when $b\equiv \pm 1(\mathrm{mod}5)$. Modulo $7$, the unit group has order $6$, soThe equation $b^4=1$ in the cyclic group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ has $\gcd (4,6)=2$ solutions, namely $b\equiv \pm 1(\mathrm{mod}7)$.

Combining the two independent sign choices by the Chinese remainder theorem givesand hence

The Chinese remainder theorem says that if the positive integers $m_1,\dots ,m_r$ are pairwise coprime and $M={\prod }_i{m}_i$, then for arbitrary residue classes $a_i(\mathrm{mod}{m}_i)$ there is a unique residue class $x(\mathrm{mod}M)$ satisfyingEquivalently, reduction is a ring isomorphism

For existence, put $M_i=M/m_i$. Pairwise coprimality gives $\gcd (M_i,m_i)=1$, so Bézout's identity supplies an integer $u_i$ with $u_i{M}_i\equiv 1(\mathrm{mod}{m}_i)$. Thenhas the required residues, because every term except the $i$th is divisible by $m_i$. If $x$ and $y$ are two solutions, each $m_i$ divides $x-y$. Pairwise coprimality then implies that their product $M$ divides $x-y$, proving uniqueness modulo $M$.

Schur lemma over $\mathbb{C}$ states that an intertwining operator $T:V\to W$ between finite-dimensional irreducible complex representations is either zero or an isomorphism. In particular, every intertwining endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility gives $\ker T=0$ and $\mathrm{im}T=W$, so $T$ is an isomorphism. For an endomorphism $T:V\to V$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. Then $T-\lambda I$ is an intertwiner with nonzero kernel, so the first part forces $T-\lambda I=0$. Hence $T=\lambda I$, completing the Proof of Schur lemma.

The actions on the ordered pair $(x,y)$ satisfyThus every element of $G$ has the form ${\sigma }^i$ or $\tau {\sigma }^i$ for $0\le i<4$, so $|G|\le 8$. Their actions areBecause $x,y$ are algebraically independent elements over $\mathbb{Q}$, these eight substitutions are distinct. Henceand the presentation identifies $G$ with the dihedral group $D_8$.

PutBoth generators $\sigma$ and $\tau$ fix $u$ and $v$, so for $F=\mathbb{Q}(u,v)$,The elements $x^2$ and $y^2$ are the two roots ofso $[F(x^2,y^2):F]\le 2$. Adjoining first $x$ and then $y$ takes at most two quadratic extensions, and the tower law givesPart (b) gives $8=|G|\le [K:K^G]$. Since $F\subseteq K^G\subseteq K$, another use of the tower law givesAll inequalities are equalities and $[K^G:F]=1$. Thereforethe dihedral fixed field of a two-variable rational function field.